C++博客-Icyflame-随笔分类-解题报告

POJ 3352 桥

Icyflame — Sun, 05 Jul 2009 08:08:00 GMT

一、题目描述

Description

It's almost summer time, and that means that it's almost summer construction time! This year, the good people who are in charge of the roads on the tropical island paradise of Remote Island would like to repair and upgrade the various roads that lead between the various tourist attractions on the island.

The roads themselves are also rather interesting. Due to the strange customs of the island, the roads are arranged so that they never meet at intersections, but rather pass over or under each other using bridges and tunnels. In this way, each road runs between two specific tourist attractions, so that the tourists do not become irreparably lost.

Unfortunately, given the nature of the repairs and upgrades needed on each road, when the construction company works on a particular road, it is unusable in either direction. This could cause a problem if it becomes impossible to travel between two tourist attractions, even if the construction company works on only one road at any particular time.

So, the Road Department of Remote Island has decided to call upon your consulting services to help remedy this problem. It has been decided that new roads will have to be built between the various attractions in such a way that in the final configuration, if any one road is undergoing construction, it would still be possible to travel between any two tourist attractions using the remaining roads. Your task is to find the minimum number of new roads necessary.

Input

The first line of input will consist of positive integers n and r, separated by a space, where 3 ≤ n ≤ 1000 is the number of tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of roads. The tourist attractions are conveniently labelled from 1 to n. Each of the following r lines will consist of two integers, v and w, separated by a space, indicating that a road exists between the attractions labelled v and w. Note that you may travel in either direction down each road, and any pair of tourist attractions will have at most one road directly between them. Also, you are assured that in the current configuration, it is possible to travel between any two tourist attractions.

Output

One line, consisting of an integer, which gives the minimum number of roads that we need to add.

Sample Input

Sample Input 1
10 12
1 2
1 3
1 4
2 5
2 6
5 6
3 7
3 8
7 8
4 9
4 10
9 10
Sample Input 2
3 3
1 2
2 3
1 3

Sample Output

Output for Sample Input 1
2
Output for Sample Input 2
0

二、分析
用DFS解决问题，详细算法：割点与桥。
三、代码

1#include<iostream>
2#include<list>
3using namespace std;
4int n, r;
5list<int> g[1001];
6int num, lab[1001], low[1001];
7list<pair<int, int> > edge;
8int degree[1001];
9int parent[1001], rank[1001];
10void init_set()
11{
12    for(int i=1; i<1001; i++)
13    {
14        parent[i] = i;
15        rank[i] = 1;
16    }
17}
18int find(int k)
19{
20    if(parent[k] == k)
21        return k;
22    else
23        return parent[k] = find(parent[k]);
24}
25void union_set(int u, int v)
26{
27    int pu = find(u), pv = find(v);
28    if(rank[pu] <= rank[pv])
29    {
30        parent[pu] = pv;
31        rank[pv] += pu;
32    }
33    else
34    {
35        parent[pv] = pu;
36        rank[pu] += pv;
37    }
38}
39void dfs(int u, int p)
40{
41    lab[u] = low[u] = num++;
42    list<int>::iterator it;
43    for(it = g[u].begin(); it != g[u].end(); it++)
44    {
45        int v = *it;
46        if(lab[v] == 0)
47        {
48            dfs(v, u);
49            low[u] = min(low[u], low[v]);
50            if(low[v] > lab[u])
51                edge.push_back(make_pair(u, v));
52            else
53                union_set(u, v); //u与v能进行缩点
54        }
55        else if(v != p)
56            low[u] = min(low[u], lab[v]);
57    }
58}
59int main()
60{
61    scanf("%d%d", &n, &r);
62    for(int i=1; i<=n; i++)
63        g[i].clear();
64    for(int i=1; i<=r; i++)
65    {
66        int v1, v2;
67        scanf("%d%d", &v1, &v2);
68        g[v1].push_back(v2);
69        g[v2].push_back(v1);
70    }
71    memset(lab, 0, sizeof lab);
72    memset(low, 0x7f, sizeof low);
73    num = 1;
74    init_set();
75    dfs(1, 0);
76    memset(degree, 0, sizeof degree);
77    list<pair<int, int> >::iterator it;
78    int res = 0;
79    for(it = edge.begin(); it != edge.end(); it++)
80    {
81        int u = it->first, v = it->second;
82        degree[find(u)]++;
83        if(degree[find(u)] == 1)
84            res++;
85        else if(degree[find(u)] == 2)
86            res--;
87        degree[find(v)]++;
88        if(degree[find(v)] == 1)
89            res++;
90        else if(degree[find(v)] == 2)
91            res--;
92    }
93    printf("%d\n", (res+1) / 2);
94}

Icyflame 2009-07-05 16:08 发表评论

POJ 1523 割点

Icyflame — Sat, 04 Jul 2009 08:12:00 GMT

一、题目描述

Description

Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible.

Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate.

Input

The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

Output

For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist.

The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.

Sample Input

Sample Output

Network #1
SPF node 3 leaves 2 subnets
Network #2
No SPF nodes
Network #3
SPF node 2 leaves 2 subnets
SPF node 3 leaves 2 subnets

二、分析
用DFS解决问题，详细算法：割点与桥。
三、代码

1#include<iostream>
2#include<list>
3using namespace std;
4int t;
5int v1, v2;
6list<int> g[1001];
7bool flag;
8int root;
9int counter;
10bool spf[1001];
11int low[1001], lab[1001];
12bool visit[1001];
13void dfs(int u, int fa)
14{
15    low[u] = lab[u] = counter++;
16    list<int>::iterator it;
17    int counter = 0;
18    for(it = g[u].begin(); it != g[u].end(); it++)
19    {
20        int v = *it;
21        if(!lab[v])
22        {
23            counter++;
24            dfs(v, u);
25            low[u] = min(low[u], low[v]);
26            if((u==root && counter>=2) || (u!=root && low[v]>=lab[u]))
27                spf[u] = flag = true;
28        }
29        else if(v != fa)
30            low[u] = min(low[u], lab[v]);
31    }
32}
33void find(int u)
34{
35    visit[u] = true;
36    list<int>::iterator it;
37    for(it = g[u].begin(); it != g[u].end(); it++)
38        if(!visit[*it])
39            find(*it);
40}
41int main()
42{
43    t = 1;
44    while(1)
45    {
46        scanf("%d", &v1);
47        if(v1 == 0) break;
48        for(int i=1; i<=1000; i++)
49            g[i].clear();
50        memset(spf, 0, sizeof spf);
51        memset(low, 0, sizeof low);
52        memset(lab, 0, sizeof lab);
53        while(v1 != 0)
54        {
55            scanf("%d", &v2);
56            g[v1].push_back(v2);
57            g[v2].push_back(v1);
58            root = v1;
59            scanf("%d", &v1);
60        }
61        counter = 1;
62        flag = false;
63        dfs(root, -1);
64        printf("Network #%d\n", t++);
65        if(flag)
66        {
67            for(int i=1; i<=1000; i++)
68            {
69                if(!spf[i]) continue;
70                int cnt = 0;
71                memset(visit, 0, sizeof visit);
72                visit[i] = true;
73                list<int>::iterator it;
74                for(it = g[i].begin(); it != g[i].end(); it++)
75                    if(!visit[*it])
76                    {
77                        cnt++;
78                        find(*it);
79                    }
80                    printf("  SPF node %d leaves %d subnets\n", i, cnt);
81            }
82        }
83        else
84            printf("  No SPF nodes\n");
85        printf("\n");
86    }
87}

Icyflame 2009-07-04 16:12 发表评论

HDOJ 2586 LCA:Tarjan

Icyflame — Thu, 02 Jul 2009 14:39:00 GMT

一、题目描述

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0 Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

Sample Input

Sample Output

二、分析
用Tarjan解决的LCA问题，详细算法：LCA问题。
三、代码

1#include<iostream>
2#include<list>
3using namespace std;
4struct node
5{
6    int v, d;
7    void init(int vv, int dd) {v = vv; d = dd;};
8};
9int t, n, m, v1, v2, len;
10list<node> g[40001];
11list<node> query[40001];
12int dis[40001];
13int res[201][3];
14int parent[40001];
15bool visit[40001];
16int find(int k)
17{
18    if(parent[k] == k)
19        return k;
20    return parent[k] = find(parent[k]);
21}
22void tarjan(int u)
23{
24    if(visit[u]) return;
25    visit[u] = true;
26    parent[u] = u;
27    list<node>::iterator it = query[u].begin();
28    while(it != query[u].end())
29    {
30        if(visit[it->v])
31            res[it->d][2] = find(it->v);
32        it++;
33    }
34    it = g[u].begin();
35    while(it != g[u].end())
36    {
37        if(!visit[it->v])
38        {
39            dis[it->v] = min(dis[it->v], dis[u] + it->d);
40            tarjan(it->v);
41            parent[it->v] = u;
42        }
43        it++;
44    }
45}
46int main()
47{
48    scanf("%d", &t);
49    while(t--)
50    {
51        scanf("%d%d", &n, &m);
52        for(int i=1; i<=n; i++)
53            g[i].clear();
54        for(int i=1; i<=m; i++)
55            query[i].clear();
56        for(int i=1; i<n; i++)
57        {
58            scanf("%d%d%d", &v1, &v2, &len);
59            node n1; n1.init(v2, len);
60            g[v1].push_back(n1);
61            node n2; n2.init(v1, len);
62            g[v2].push_back(n2);
63        }
64        for(int i=1; i<=m; i++)
65        {
66            scanf("%d%d", &v1, &v2);
67            res[i][0] = v1;
68            res[i][1] = v2;
69            node n1; n1.init(v2, i);
70            query[v1].push_back(n1);
71            node n2; n2.init(v1, i);
72            query[v2].push_back(n2);
73        }
74        memset(visit, 0, sizeof(visit));
75        memset(dis, 0x7f, sizeof(dis));
76        dis[1] = 0;
77        tarjan(1);
78        for(int i=1; i<=m; i++)
79            printf("%d\n", dis[res[i][0]] + dis[res[i][1]] - 2*dis[res[i][2]]);
80    }
81}

Icyflame 2009-07-02 22:39 发表评论

ZOJ 3195 LCA:RMQ

Icyflame — Thu, 02 Jul 2009 12:55:00 GMT

摘要: 一、题目描述 Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of the... 阅读全文

Icyflame 2009-07-02 20:55 发表评论

POJ 3368 RMQ

Icyflame — Wed, 01 Jul 2009 08:24:00 GMT

摘要: 一、题目描述 Description You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ... 阅读全文

Icyflame 2009-07-01 16:24 发表评论

POJ 2516 最小费用最大流

Icyflame — Tue, 30 Jun 2009 14:09:00 GMT

摘要: 一、题目描述 Description Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy h... 阅读全文

Icyflame 2009-06-30 22:09 发表评论

POJ 2195 最优匹配

Icyflame — Mon, 29 Jun 2009 06:42:00 GMT

摘要: 一、题目描述 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each l... 阅读全文

Icyflame 2009-06-29 14:42 发表评论

POJ 1149 最大流

Icyflame — Fri, 26 Jun 2009 08:15:00 GMT

摘要: 一、题目描述 Description Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. ... 阅读全文

Icyflame 2009-06-26 16:15 发表评论

POJ 1459 最大流

Icyflame — Wed, 24 Jun 2009 11:38:00 GMT

一、题目描述

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c_max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
(3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
(0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

二、分析
增加点n为s，点n+1为t，求最大流，使用Push-Relabel算法，具体算法：最大流问题
三、代码

1#include<iostream>
2using namespace std;
3#define MAXN 202
4int s, t;
5int n, np, nc, m;
6char str[50];
7int c[MAXN][MAXN];
8int f[MAXN][MAXN];
9int e[MAXN];
10int h[MAXN];
11void push(int u, int v)
12{
13    int d = min(e[u], c[u][v] - f[u][v]);
14    f[u][v] += d;
15    f[v][u] = -f[u][v];
16    e[u] -= d;
17    e[v] += d;
18}
19bool relabel(int u)
20{
21    int mh = INT_MAX;
22    for(int i=0; i<n+2; i++)
23    {
24        if(c[u][i] > f[u][i])
25            mh = min(mh, h[i]);
26    }
27    if(mh == INT_MAX)
28        return false; //残留网络中无从u出发的路
29    h[u] = mh + 1;
30    for(int i=0; i<n+2; i++)
31    {
32        if(e[u] == 0) //已无余流，不需再次push
33            break;
34        if(h[i] == mh && c[u][i] > f[u][i]) //push的条件
35            push(u, i);
36    }
37    return true;
38}
39void init_preflow()
40{
41    memset(h, 0, sizeof(h));
42    memset(e, 0, sizeof(e));
43    h[s] = n+2;
44    for(int i=0; i<n+2; i++)
45    {
46        if(c[s][i] == 0)
47            continue;
48        f[s][i] = c[s][i];
49        f[i][s] = -f[s][i];
50        e[i] = c[s][i];
51        e[s] -= c[s][i];
52    }
53}
54void push_relabel()
55{
56    init_preflow();
57    bool flag = true; //表示是否还有relabel操作
58    while(flag)
59    {
60        flag = false;
61        for(int i=0; i<n; i++)
62            if(e[i] > 0)
63                flag = flag || relabel(i);
64    }
65}
66int main()
67{
68    while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF)
69    {
70        s = n; t = n+1;
71        memset(c, 0, sizeof(c));
72        memset(f, 0, sizeof(f));
73        while(m--)
74        {
75            scanf("%s", &str);
76            int u=0, v=0, z=0;
77            sscanf(str, "(%d,%d)%d", &u, &v, &z);
78            c[u][v] = z;
79        }
80        for(int i=0; i<np+nc; i++)
81        {
82            scanf("%s", &str);
83            int u=0, z=0;
84            sscanf(str, "(%d)%d", &u, &z);
85            if(i < np)
86                c[s][u] = z;
87            else if(i >= np && i < np + nc)
88                c[u][t] = z;
89        }
90        push_relabel();
91        printf("%d\n", e[t]);
92    }
93}

Icyflame 2009-06-24 19:38 发表评论

POJ 1273 最大流

Icyflame — Tue, 23 Jun 2009 11:38:00 GMT

一、题目描述

Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

Sample Output

二、分析
其实就是以点1为s，以点m为t，求最大流，但是要注意输入的路径可以重复(见代码30行)，使用Edmonds-Karp算法，具体算法：最大流问题
三、代码

1#include<iostream>
2#include<queue>
3using namespace std;
4#define MAXM 201
5int m, n;
6int si, ei, ci;
7int c[MAXM][MAXM];
8int f[MAXM][MAXM];
9int cf[MAXM][MAXM];
10bool visit[MAXM];
11int p[MAXM];
12struct node
13{
14    int v, cf;
15    void set(int vv, int ccf)
16    {
17        v = vv; cf = ccf;
18    }
19};
20int main()
21{
22    while(scanf("%d%d", &n, &m) != EOF)
23    {
24        memset(c, 0, sizeof(c));
25        memset(f, 0, sizeof(f));
26        memset(cf, 0, sizeof(cf));
27        while(n--)
28        {
29            scanf("%d%d%d", &si, &ei, &ci);
30            c[si][ei] += ci;
31            cf[si][ei] = c[si][ei];
32        }
33        bool flag = true; //用于表示是否找到增广路
34        while(flag)
35        {
36            flag = false;
37            memset(visit, 0, sizeof(visit));
38            queue<node> q;
39            node temp;
40            temp.set(1, INT_MAX);
41            p[1] = 0;
42            q.push(temp); visit[1] = true;
43            while(!q.empty()) //广度优先搜索
44            {
45                node temp = q.front(); q.pop();
46                for(int i=1; i<=m; i++)
47                {
48                    if(temp.v == i || visit[i] || cf[temp.v][i] == 0)
49                        continue;
50                    node newNode;
51                    newNode.set(i, min(temp.cf, cf[temp.v][i]));
52                    p[i] = temp.v;
53                    q.push(newNode);
54                    visit[i] = true;
55                    if(i == m)
56                    {
57                        flag = true; //找到增广路
58                        break;
59                    }
60                }
61                if(flag)
62                    break;
63            }
64            if(flag)
65            {
66                int mincf = q.back().cf;
67                int v1 = p[m], v2 = m;
68                while(v1 != 0)
69                {
70                    f[v1][v2] += mincf; //修改流
71                    f[v2][v1] = -f[v1][v2];
72                    cf[v1][v2] = c[v1][v2] - f[v1][v2]; //修改残留容量
73                    cf[v2][v1] = c[v2][v1] - f[v2][v1];
74                    v2 = v1;
75                    v1 = p[v1];
76                }
77            }
78        }
79        int res = 0;
80        for(int i=2; i<=m; i++) //计算最大流
81            res += f[1][i];
82        printf("%d\n", res);
83    }
84}

Icyflame 2009-06-23 19:38 发表评论