C++博客-算法的力量-随笔分类-算法与数据结构

最大流 Edmonds-Karp

reiks — Sat, 29 Aug 2009 05:39:00 GMT

//Edmonds-Karp
//return the largest flow;flow[] will record every edge's flow
//n, the number of nodes in the graph;cap, the capacity
//O(VE^2)
#define N 100
#define inf 0x3f3f3f3f
int Edmonds_Karp(int n,int cap[][N],int source,int sink)
{
    int flow[N][N];
    int pre[N],que[N],d[N]; // d 是增广路长度，pre 记录前驱，que是BFS队列
    int p,q,t,i,j;
    if (source==sink) return inf;
    memset(flow,0,sizeof(flow));
    while (true)
    {
        memset(pre,-1,sizeof(pre));
        d[source]=inf;
        p=q=0, que[q++] = source;
        while(p < q&&pre[sink]<0)    // BFS 找路径
        {
            t=que[p++];
            for (i=0;i<n;i++)
                if ( pre[i]<0 && (j=cap[t][i]-flow[t][i]) ) // j取得残余路径值
                    pre[que[q++] = i] = t,d[i] = min(d[t], j);
        }
        if (pre[sink]<0) break;    // 找不到增广路，退出
        for (i=sink; i!=source; i=pre[i])
        {
            flow[pre[i]][i]+=d[sink];    // 正向流量加
            flow[i][pre[i]]-=d[sink];    // 反向流量减
        }
    }
    for (j=i=0; i<n; j+=flow[source][i++]);
    return j;
}

reiks 2009-08-29 13:39 发表评论

匈牙利算法

reiks — Fri, 28 Aug 2009 02:35:00 GMT

摘要: #includebool map[102][302],use[302];int link[302],n,m;bool dfs(int);int main(){ int t,v,i,j,x,num; scanf("%d",&am... 阅读全文

reiks 2009-08-28 10:35 发表评论

树状数组

reiks — Fri, 28 Aug 2009 02:35:00 GMT

#include <stdio.h>
#include <memory.h>
#define N 1000

class treearray
{
       public:
       int c[N],n;
       void clear()
       {
             memset(this,0,sizeof(*this));
       }
       int lowbit(int x)
       {
             return x&(x^(x-1));
       }
       void change(int i,int d)
       {
             for (;i<=n;i+=lowbit(i))       c[i]+=d;
       }
       int getsum(int i)
       {
             int t;
             for (t=0;i>0;i-=lowbit(i))       t+=c[i];
             return t;
       }
}t;

main()//附一个测试程序
{
       int i,x;
       t.clear();
       scanf("%d",&t.n);
       for (i=1;i<=t.n;i++)
       {
             scanf("%d",&x);
             t.change(i,x);
       }
       for (;scanf("%d",&x),x;)       printf("%d\n",t.getsum(x));
       return 0;
}

reiks 2009-08-28 10:35 发表评论

并查集

reiks — Fri, 28 Aug 2009 02:34:00 GMT

#define MAXSIZE 50001
int father[MAXSIZE];
int rank[MAXSIZE];

void initial()
{
    memset(rank, 0, sizeof(rank));
    for ( int i = 0; i < MAXSIZE; ++i )
        father[i] = -1;
}

int find_set(int x)
{
    int r = x, q;

    while(father[r] != -1)
    {
        r = father[r];
    }

    while(x != r)
    {
        q = father[x];
        father[x] = r;
        x = q;
    }
    return r;
}

void union_set(int x, int y)
{
    int a = find_set(x);
    int b = find_set(y);
    if (a == b)
        return;
    if (rank[a] > rank[b])
    {
        father[b] = a;
    }
    else
    {
        father[a] = b;
        if (rank[a] == rank[b])
        {
            ++rank[b];
        }
    }
}

reiks 2009-08-28 10:34 发表评论

Topsort

reiks — Fri, 28 Aug 2009 02:33:00 GMT

/**
* TOPSORT(简单版) 拓扑排序(Topological Sort)
* 输入：有向图g
* 输出：是否存在拓扑排序，如果存在，获取拓扑排序序列seq
* 结构：图g用邻接矩阵表示
* 算法：广度优先搜索(BFS)
* 复杂度：O(|V|^2)
*/

#include <iostream>
#include <vector>
#include <queue>
#include <iterator>
#include <algorithm>
#include <numeric>
#include <climits>
using namespace std;

int n;                            // n ：顶点个数
vector<vector<int> > g;           // g ：图(graph)(用邻接矩阵(adjacent matrix)表示)
vector<int> seq;                // seq ：拓扑序列(sequence)

bool TopSort()
{
    vector<int> inc(n, 0);
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
             if (g[i][j] < INT_MAX) ++inc[j]; // 计算每个顶点的入度，
    queue<int> que;
    for (int j = 0; j < n; ++j)
        if (inc[j] == 0) que.push(j); // 如果顶点的入度为0，入队。
    int seqc = 0;
    seq.resize(n);
    while (!que.empty())     // 如果队列que非空，
    {
        int v = que.front(); que.pop();
        seq[seqc++] = v;      // 顶点v出队，放入seq中，
        for (int w = 0; w < n; ++w)     // 遍历所有v指向的顶点w，
            if (g[v][w] < INT_MAX)
                if (--inc[w] == 0) que.push(w); // 调整w的入度，如果w的入度为0，入队。
    }
    return seqc == n; // 如果seq已处理顶点数为n，存在拓扑排序，否则存在回路。
}

int main()
{
    n = 7;
    g.assign(n, vector<int>(n, INT_MAX));
    g[0][1] = 1, g[0][2] = 1, g[0][3] = 1;
    g[1][3] = 1, g[1][4] = 1;
    g[2][5] = 1;
    g[3][2] = 1, g[3][5] = 1, g[3][6] = 1;
    g[4][3] = 1, g[4][6] = 1;
    g[6][5] = 1;

    if (TopSort())
    {
         copy(seq.begin(), seq.end(), ostream_iterator<int>(cout, " "));
         cout << endl;
    }
    else
    {
         cout << "circles exist" << endl;
    }

    system("pause");
    return 0;
}

reiks 2009-08-28 10:33 发表评论

Trie树

reiks — Fri, 28 Aug 2009 02:32:00 GMT

/*
Name: Trie树的基本实现
Author: MaiK
Description: Trie树的基本实现 ,包括查找插入和删除操作(卫星数据可以因情况而异)
*/
#include<algorithm>
#include<iostream>
using namespace std;

const int sonnum=26,base='a';
struct Trie
{
    int num;  //to remember how many word can reach here,that is to say,prefix
    bool terminal;  //If terminal==true ,the current point has no following point
    struct Trie *son[sonnum];  //the following point
};
Trie *NewTrie()// create a new node
{
    Trie *temp=new Trie;
    temp->num=1;
    temp->terminal=false;
    for (int i=0; i<sonnum; ++i)
        temp->son[i] = NULL;
    return temp;
}
void Insert(Trie *pnt,char *s,int len)// insert a new word to Trie tree
{
    Trie *temp=pnt;
    for (int i=0;i<len;++i)
    {
        if (temp->son[s[i]-base]==NULL)
            temp->son[s[i]-base]=NewTrie();
        else
            temp->son[s[i]-base]->num++;
        temp=temp->son[s[i]-base];
    }
    temp->terminal=true;
}
void Delete(Trie *pnt)  // delete the whole tree
{
    if (pnt!=NULL)
    {
        for (int i=0;i<sonnum;++i)
            if (pnt->son[i]!=NULL)
                Delete(pnt->son[i]);
        delete pnt;
        pnt=NULL;
    }
}
Trie* Find(Trie *pnt,char *s,int len)  //trie to find the current word
{
    Trie *temp=pnt;
    for (int i=0;i<len;++i)
        if (temp->son[s[i]-base]!=NULL)
            temp=temp->son[s[i]-base];
        else return NULL;
    return temp;
}

reiks 2009-08-28 10:32 发表评论

大整数乘除小整数

reiks — Fri, 28 Aug 2009 01:22:00 GMT

// 大整数乘以一个小整数
void big_mul(int d[], int s[], int n)
{
    int plus = 0;
    for (int i = 1; i < 61; ++i)
    {
        d[i] = s[i] * n;
        d[i] += plus;
        plus = d[i] / 10;
        d[i] %= 10;
    }
}

// 大整数除以一个小整数
void big_div(int d[], int s[], int n)
{
    int left = 0;
    for (int i = 60; i > 0; --i)
    {
        left *= 10;
        left += s[i];
        if (left < n)
        {
            d[i] = 0;
        }
        else
        {
            d[i] = left / n;
            left %= n;
        }
    }
}

reiks 2009-08-28 09:22 发表评论

RMQ问题ST算法

reiks — Fri, 28 Aug 2009 01:20:00 GMT

/*
RMQ(Range Minimum/Maximum Query)问题：
   RMQ问题是求给定区间中的最值问题。当然，最简单的算法是O(n)的，但是对于查询次数很多（设置多大100万次），O(n)的算法效率不够。可以用线段树将算法优化到O（logn)（在线段树中保存线段的最值）。不过，Sparse_Table算法才是最好的：它可以在O(nlogn)的预处理以后实现O(1)的查询效率。下面把Sparse Table算法分成预处理和查询两部分来说明(以求最小值为例)。

预处理:
预处理使用DP的思想，f(i, j)表示[i, i+2^j - 1]区间中的最小值，我们可以开辟一个数组专门来保存f(i, j)的值。
例如，f(0, 0)表示[0,0]之间的最小值,就是num[0], f(0, 2)表示[0, 3]之间的最小值, f(2, 4)表示[2, 17]之间的最小值
注意, 因为f(i, j)可以由f(i, j - 1)和f(i+2^(j-1), j-1)导出, 而递推的初值(所有的f(i, 0) = i)都是已知的
所以我们可以采用自底向上的算法递推地给出所有符合条件的f(i, j)的值。

查询:
假设要查询从m到n这一段的最小值, 那么我们先求出一个最大的k, 使得k满足2^k <= (n - m + 1).
于是我们就可以把[m, n]分成两个(部分重叠的)长度为2^k的区间: [m, m+2^k-1], [n-2^k+1, n];
而我们之前已经求出了f(m, k)为[m, m+2^k-1]的最小值, f(n-2^k+1, k)为[n-2^k+1, n]的最小值
我们只要返回其中更小的那个, 就是我们想要的答案, 这个算法的时间复杂度是O(1)的.
例如, rmq(0, 11) = min(f(0, 3), f(4, 3))
*/

#include<iostream>
#include<cmath>
using namespace std;
#define MAXN 1000000
#define mmin(a, b)   ((a)<=(b)?(a):(b))
#define mmax(a, b)   ((a)>=(b)?(a):(b))

int num[MAXN];
int f1[MAXN][100];
int f2[MAXN][100];

//测试输出所有的f(i, j)
void dump(int n)
{
    int i, j;
    for(i = 0; i < n; i++)
    {
        for(j = 0; i + (1<<j) - 1 < n; j++)
        {
            printf("f[%d, %d] = %d\t", i, j, f1[i][j]);
        }
        printf("\n");
    }
    for(i = 0; i < n; i++)
       printf("%d ", num[i]);
    printf("\n");
    for(i = 0; i < n; i++)
    {
        for(j = 0; i + (1<<j) - 1 < n; j++)
        {
            printf("f[%d, %d] = %d\t", i, j, f2[i][j]);
        }
        printf("\n");
    }
    for(i = 0; i < n; i++)
        printf("%d ", num[i]);
    printf("\n");
}

//sparse table算法
void st(int n)
{
    int i, j, k, m;
    k = (int) (log((double)n) / log(2.0));
    for(i = 0; i < n; i++)
    {
        f1[i][0] = num[i]; //递推的初值
        f2[i][0] = num[i];
    }
    for(j = 1; j <= k; j++)
    { //自底向上递推
        for(i = 0; i + (1 << j) - 1 < n; i++)
        {
            m = i + (1 << (j - 1)); //求出中间的那个值
            f1[i][j] = mmax(f1[i][j-1], f1[m][j-1]);
            f2[i][j] = mmin(f2[i][j-1], f2[m][j-1]);
        }
    }
}

//查询i和j之间的最值,注意i是从0开始的
void rmq(int i, int j)
{
    int k = (int)(log(double(j-i+1)) / log(2.0)), t1, t2; //用对2去对数的方法求出k
    t1 = mmax(f1[i][k], f1[j - (1<<k) + 1][k]);
    t2 = mmin(f2[i][k], f2[j - (1<<k) + 1][k]);
    printf("%d\n",t1 - t2);
}

int main()
{
    int i,N,Q,A,B;
    scanf("%d %d", &N, &Q);
    for (i = 0; i < N; ++i)
    {
        scanf("%d", num+i);
    }

    st(N); //初始化
    //dump(N); //测试输出所有f(i, j)
    while(Q--)
    {
        scanf("%d %d",&A,&B);
        rmq(A-1, B-1);
    }
    return 0;
}

reiks 2009-08-28 09:20 发表评论

MST-Kruskal

reiks — Fri, 28 Aug 2009 01:17:00 GMT

#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

struct Edge
{
    int x, y, w; /*边集，边(x,y)，权为c*/
}e[20001];

int rank[1001]; /*节点的秩*/
int p[1001]; /*p[x]表示x节点父节点*/
int ans=0;
int ma;

void Make(int x)
{
    p[x] = x;
    rank[x] = 1;
}

int Find(int x) /*查找x所在集合子树的根*/
{
    if (x == p[x]) return x;
    return p[x] = Find( p[x] );;
}

void Union(int x, int y, int c)
{
    x = Find(x);
    y = Find(y);
    if ( x != y ) /*若x,y不属于同一集合*/
    {
        if ( rank[x] > rank[y] ) /*将秩较小的树连接到秩较大的树后*/
        {
            p[y] = x;
        }
        else
        {
            if(rank[x] == rank[y])
                rank[y]++;
            p[x] = y;
        }
        ans += c;
        ma++;
    }
}

bool cmp (const Edge & a, const Edge & b)
{
    return a.w > b.w;
}

int N, M;

int main()
{
    int n; /*边的条数*/
    int i;
    //ifstream in("123.txt");
    ans=0;
    ma = 1;
    scanf("%d %d", &N, &M);
    //cin >> N >> M;
    for ( i = 1; i <= N; ++i)
        Make(i);
    for ( i = 0; i < M; ++i)
    {
        scanf("%d %d %d", &e[i].x, &e[i].y, &e[i].w);
    }

    sort(e, e + M, cmp); /*按权值非降序排序*/

    for ( i = 0; i < M; ++i)
    {
        Union(e[i].x, e[i].y, e[i].w);
    }
    if (ma == N)
        printf("%d", ans);
    else
        printf("-1");
    //system("pause");
    return 0;
}

reiks 2009-08-28 09:17 发表评论

组合数

reiks — Fri, 28 Aug 2009 01:16:00 GMT

#include <stdio.h>
#include <assert.h>
#include <malloc.h>

void c(int n)
{
    extern void _c(int n, int cur, int *a, int now);
    int *a;
    a = (int *) malloc(n * sizeof(int));
    assert(a != NULL);
    _c(n, 0, a, 1);
    free(a);
}

void _c(int n, int cur, int *a, int now)
{
    int i,j;
    for (i=now; i<=n; i++)
    {
        a[cur] = i;

        for (j=0; j<=cur; j++)
        {
            printf("%d ", a[j]);
        }
        printf("\n");

        _c(n, cur+1, a, i+1);
    }
}

int main()
{
    c(3);
}

reiks 2009-08-28 09:16 发表评论

全排列

reiks — Fri, 28 Aug 2009 01:15:00 GMT

#include <stdio.h>
#include <assert.h>
#include <malloc.h>

void p(int n)
{
    extern void _p(int n, int cur, int *a);
    int *a;
    a = (int *) malloc(n * sizeof(int));
    assert(a != NULL);
    _p(n, 0, a);
    free(a);
}

void _p(int n, int cur, int *a)
{
    int i,j;
    if (cur == n)
    {
        for (j=0; j<cur; j++)
        {
            printf("%d ", a[j]);
        }
        printf("\n");
        return;
    }
    for (i=1; i<=n; i++)
    {
        for (j=0; j<cur; j++)
        {
            if (a[j] == i)
            {
                break;
            }
        }
        if (j>=cur)
        {
            a[cur] = i;
            _p(n, cur+1, a);
        }
    }
}

int main()
{
    p(3);
}

reiks 2009-08-28 09:15 发表评论

Dijkstra

reiks — Fri, 28 Aug 2009 01:14:00 GMT

#include <iostream>
#include <vector>
#include <list>
#include <iterator>
#include <algorithm>
#include <numeric>
#include <functional>
#include <climits>
using namespace std;

int n;                    // n : 顶点个数
vector<vector<int> > g; // g : 图(graph)(用邻接矩阵(adjacent matrix)表示)
int s;                    // s : 源点(source)
vector<bool> known;        // known : 各点是否知道最短路径
vector<int> dist;        // dist : 源点s到各点的最短路径长
vector<int> prev;        // prev : 各点最短路径的前一顶点

void Dijkstra()            // 贪心算法(Greedy Algorithm)
{
    known.assign(n, false);
    dist.assign(n, INT_MAX);
    prev.resize(n);            // 初始化known、dist、prev。
    dist[s] = 0;            // 初始化源点s到自身的路径长为0。
    for (;;)
    {
        int min = INT_MAX, v = s;
        for (int i = 0; i < n; ++i)
            if (!known[i] && min > dist[i])
                min = dist[i], v = i;    // 寻找未知的最短路径长的顶点v，
        if (min == INT_MAX) break;        // 如果找不到，退出；
        known[v] = true;                // 如果找到，将顶点v设为已知，
        for (int w = 0; w < n; ++w)        // 遍历所有v指向的顶点w，
            if (!known[w] && g[v][w] < INT_MAX && dist[w] > dist[v] + g[v][w])
                dist[w] = dist[v] + g[v][w], prev[w] = v;    // 调整顶点w的最短路径长dist和最短路径的前一顶点 prev。
    }
}

void Print_SP(int v)
{
     if (v != s) Print_SP(prev[v]);
     cout << v << " ";
}

int main()
{
    n = 7;
    g.assign(n, vector<int>(n, INT_MAX));
    g[0][1] = 2; g[0][3] = 1;
    g[1][3] = 3; g[1][4] = 10;
    g[2][0] = 4; g[2][5] = 5;
    g[3][2] = 2; g[3][4] = 2; g[3][5] = 8; g[3][6] = 4;
    g[4][6] = 6;
    g[6][5] = 1;

    s = 0;
    Dijkstra();

    copy(dist.begin(), dist.end(), ostream_iterator<int>(cout, " ")); cout << endl;
    for (int i = 0; i < n; ++i)
        if(dist[i] != INT_MAX)
        {
            cout << s << "->" << i << ": ";
            Print_SP(i);
            cout << endl;
        }

    system("pause");
    return 0;
}

/*============优先队列版================*/
class great
{
public:
        bool operator() (pair<int, int>& p1, pair<int, int>& p2) {
                return (p1.second > p2.second);
        }
};

int G[N][N];

int dijkstra(int src, int dst) {
        vector<int> cost(N, INT_MAX);
        vector<bool> visited(N, false);

        priority_queue< pair<int, int>, vector< pair<int, int> >, great > Q;

        cost[src] = 0;
        Q.push( make_pair(src, 0) );

        while(!Q.empty()) {
                pair<int, int> top = Q.top();
                Q.pop();

                int v = top.first;
                if (v == dst) return cost[v];

                if (visited[v]) continue;
                visited[v] = true;

                for(int v2 = 0; v2 < N; v2++) if (G[v][v2] != 0) {
                        int dist = G[v][v2];
                        if(cost[v2] > cost[v] + dist) {
                                cost[v2] = cost[v] + dist;
                                Q.push( make_pair(v2, cost[v2]) );
                        }
                }
        }

        return -1;
}

reiks 2009-08-28 09:14 发表评论

Floyd

reiks — Fri, 28 Aug 2009 01:13:00 GMT

#include<fstream>
#define Maxm 501
using namespace std;
ifstream fin("APSP.in");
ofstream fout("APSP.out");
int p, q, k, m;
int Vertex, Line[Maxm];
int Path[Maxm][Maxm], Map[Maxm][Maxm], Dist[Maxm][Maxm];
void Root(int p,int q)
{
    if (Path[p][q]>0)
    {
        Root(p, Path[p][q]);
        Root(Path[p][q], q);
    }
    else
    {
        Line[k]=q;
        k++;
    }
}
int main()
{
    memset(Path,0,sizeof(Path));
    memset(Map,0,sizeof(Map));
    memset(Dist,0,sizeof(Dist));
    fin >> Vertex;
    for(p=1;p<=Vertex;p++)
        for(q=1;q<=Vertex;q++)
        {
            fin >> Map[p][q];
            Dist[p][q]=Map[p][q];
        }
    for(k=1;k<=Vertex;k++)
    {
        for(p=1;p<=Vertex;p++)
        {
            if (Dist[p][k]>0)
            {
                for(q=1;q<=Vertex;q++)
                {
                    if (Dist[k][q]>0)
                    {
                        if (((Dist[p][q]>Dist[p][k]+Dist[k][q])||(Dist[p][q]==0))&&(p!=q))
                        {
                            Dist[p][q]=Dist[p][k]+Dist[k][q];
                            Path[p][q]=k;
                        }
                    }
                }
            }
        }
    }
    for(p=1;p<=Vertex;p++)
    {
        for(q=p+1;q<=Vertex;q++)
        {
            fout << "\n==========================\n";
            fout << "Source:" << p << '\n' << "Target " << q << '\n';
            fout << "Distance:" << Dist[p][q] << '\n';
            fout << "Path:" << p;
            k=2;
            Root(p,q);
            for(m=2;m<=k-1;m++)
            fout << "-->" << Line[m];
            fout << '\n';
            fout << "==========================\n";
        }
    }
    fin.close();
    fout.close();
    return 0;
}
/*
注解:无法连通的两个点之间距离为0;
Sample Input
7
00 20 50 30 00 00 00
20 00 25 00 00 70 00
50 25 00 40 25 50 00
30 00 40 00 55 00 00
00 00 25 55 00 10 70
00 70 50 00 10 00 50
00 00 00 00 70 50 00
*/

reiks 2009-08-28 09:13 发表评论

MST-PRIM

reiks — Fri, 28 Aug 2009 01:12:00 GMT

摘要: // PRIM(简单版) 最小生成树算法 (Minimum Spanning Tree) // 输入：图g； /... 阅读全文

reiks 2009-08-28 09:12 发表评论

线段树

reiks — Fri, 28 Aug 2009 01:11:00 GMT

#include <iostream>
using namespace std;
#define N 100
struct TNode
{
    int left, right;
    int n;
};
TNode T[N*2+1];
void build(int s, int t, int step)
{
    T[step].left = s;
    T[step].right = t;
    T[step].n = 0;
    if(s == t)
        return;
    int mid = (s + t) / 2;
    build(s, mid, step*2);
    build(mid+1, t, step*2+1);
}
void insert(int s, int t, int step)
// insert [s, t] in the tree
{
    if(s == T[step].left && t == T[step].right)
    {
        T[step].n++;
        return;
    }
    int mid = (T[step].left + T[step].right) / 2;
    if(t <= mid)
        insert(s, t, step*2);
    else if(s > mid)
        insert(s, t, step*2+1);
    else
    {
        insert(s,mid, step*2);
        insert(mid+1, t, step*2+1);
    }
}
void calculate(int s, int t, int step, int target, int& count)
// caculate target in the tree
{
    count += T[step].n;
    if(s == t)
        return;
    int mid = (s + t) / 2;
    if(target <= mid)
        calculate(s, mid, step*2, target, count);
    else
        calculate(mid+1, t, step*2+1, target, count);
}
int main()
{
    build(0, 7, 1);
    insert(2, 5, 1);
    insert(4, 6, 1);
    insert(0, 7, 1);
    int count = 0;
    calculate(0, 7, 1, 4, count);
    cout  << count << endl;
    return 0;
}

reiks 2009-08-28 09:11 发表评论