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HNU 11072 && PKU 3635 Full Tank

晓天_xiaotian — Fri, 08 Oct 2010 09:08:00 GMT

优先队列bfs求最短路

1 /*
2  * File:   PKU 3635 Full Tank?
3  * Author: xiaotian @ hnu
4  * Created on 2010年10月8日, 下午4:27
5  * 题解：优先队列bfs
6  */
7 #include<iostream>
8 #include<math.h>
9 #include<queue>
10 #include<stdio.h>
11 #include<algorithm>
12 #include<string.h>
13 #include<vector>
14 using namespace std;
15 #define N 1008
16 #define inf 0x7fffffff
17
18 struct node {
19     int v, dist;
20
21     node(int a = 0, int b = 0) : v(a), dist(b) {
22     }
23 };
24
25 struct Node {
26     int v,rest,cost;
27
28     Node(int a = 0, int b = 0, int c = 0) : v(a), rest(b), cost(c) {
29     }
30     bool operator<(const Node &r) const{
31         return r.cost < cost;
32     }
33 };
34 vector<node>g[N];
35 int dp[N][102];
36 bool vist[N][102];
37 int n, m, a[N], V, S, T, ans;
38 priority_queue<Node>q;
39
40 int bfs() {
41     int x, y, z;
42     while (!q.empty())
43         q.pop();
44     q.push(Node(S, 0, 0));
45     dp[S][0] = 0;
46     while (!q.empty()) {
47         Node p = q.top();
48         q.pop();
49         x = p.v, y = p.rest, z = p.cost;
50         vist[x][y] = 1;
51         if (x == T) return z;
52         if (y + 1 <= V && !vist[x][y + 1] && dp[x][y + 1] > dp[x][y] + a[x]) {
53             dp[x][y + 1] = dp[x][y] + a[x];
54             q.push(Node(x, y + 1, dp[x][y + 1]));
55         }
56         for (int i = 0; i < g[x].size(); i++) {
57             int t = g[x][i].v;
58             int k = y - g[x][i].dist;
59             if (k >= 0 && !vist[t][k] && z < dp[t][k]) {
60                 dp[t][k] = z;
61                 q.push(Node(t, k, dp[t][k]));
62             }
63         }
64     }
65     return -1;
66 }
67
68 int main() {
69     int i, j;
70     while (scanf("%d%d", &n, &m) != EOF) {
71         for (i = 0; i < n; i++) {
72             scanf("%d", &a[i]);
73             g[i].clear();
74         }
75         int u, v, w;
76         for (i = 0; i < m; i++) {
77             scanf("%d%d%d", &u, &v, &w);
78             g[u].push_back(node(v, w));
79             g[v].push_back(node(u, w));
80         }
81         int tc;
82         scanf("%d", &tc);
83         while (tc--) {
84             scanf("%d%d%d", &V, &S, &T);
85             memset(vist, 0, sizeof (vist));
86             for (i = 0; i < n; i++)
87                 for (j = 0; j <= 100; j++)
88                     dp[i][j] = inf;
89             ans = bfs();
90             if (ans == -1)
91                 printf("impossible\n");
92             else
93                 printf("%d\n", ans);
94         }
95     }
96     return 0;
97 }

晓天_xiaotian 2010-10-08 17:08 发表评论

Timus 1040. Airline company

晓天_xiaotian — Sat, 02 Oct 2010 02:13:00 GMT

/*
* File: Timus 1040. Airline company
* Author: xiaotian @ hnu
* Created on 2010年10月2日, 上午9:34
* 题解：一次找极长路（即两边不能再加边），将他们依次编号，然后删掉这些边。直至图中没有边为止。输出解即可。
* 不会出现没有解的情况。
*/

1 #include <iostream>
2 #include<string.h>
3 #include<iostream>
4 #define N 55
5 using namespace std;
6 int n, m;
7 int g[N][N];
8 int link[N][N];
9 int num[N*N];
10 int now;
11 bool vis[N];
12 bool p[N*N];
13
14 void dfs(int x) {
15     for (int i = 1; i <= g[x][0]; ++i)
16         if (!p[link[x][i]]) {
17             p[link[x][i]] = true;
18             num[link[x][i]] = ++now;
19             if (!vis[g[x][i]]) {
20                 vis[g[x][i]] = true;
21                 dfs(g[x][i]);
22             }
23         }
24 }
25
26 int main() {
27     while (scanf("%d %d", &n, &m) != EOF) {
28         memset(vis,0,sizeof(vis));
29         memset(p,0,sizeof(p));
30         now=0;
31         for (int i = 0; i <= n; i++) g[i][0] = link[i][0] = 0;
32         for (int i = 1; i <= m; ++i) {
33             int a, b;
34             scanf("%d %d", &a, &b);
35             g[a][++g[a][0]] = b;
36             g[b][++g[b][0]] = a;
37             link[a][++link[a][0]] = i;
38             link[b][++link[b][0]] = i;
39         }
40         vis[1] = true;
41         dfs(1);
42         puts("YES");
43         for (int i = 1; i <= m; ++i) printf("%d ", num[i]);
44     }
45 }

晓天_xiaotian 2010-10-02 10:13 发表评论

Herbin 1004 Power Stations

晓天_xiaotian — Wed, 29 Sep 2010 07:27:00 GMT

摘要: 第35届ACM/ICPC亚洲区预选赛哈尔滨赛区的 1004 题，Power Stations。Dancing Links，相对于其他的 DLX 来说，这道题目应该还算简单的。 Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> 1... 阅读全文

晓天_xiaotian 2010-09-29 15:27 发表评论

Timus 1033. Labyrinth

晓天_xiaotian — Wed, 29 Sep 2010 03:23:00 GMT

1 /*
2  * File:   Timus 1033. Labyrinth
3  * Author: xiaotian @ hnu
4  * Created on 2010年9月29日, 上午10:30
5  * 题解：bfs数墙面个数，最后乘以 9 。
6  */
7
8 #include<stdio.h>
9 #include<iostream>
10 #include<string.h>
11 #include<queue>
12 #include<set>
13 #include<math.h>
14 using namespace std;
15 #define N 40
16 #define inf 0x7ffffff
17 const int dx[4] = {0, 1, 0, -1};
18 const int dy[4] = {1, 0, -1, 0};
19 char map[N][N];
20 int vis[N][N][4];
21 int num[N][N];
22 int n;
23
24 struct point {
25     int x, y;
26     point(int xx = 0, int yy = 0) : x(xx), y(yy) {
27     }
28 };
29
30 inline int ok(int x, int y) {
31     return x > 0 && x <= n && y > 0 && y <= n;
32 }
33
34 int bfs(point s) {
35     int ans = 0;
36     queue<point>q;
37     q.push(s);
38     num[s.x][s.y] = 1;
39     while (!q.empty()) {
40         point p = q.front();
41         q.pop();
42         for (int i = 0; i < 4; i++) {
43             int x = p.x + dx[i], y = p.y + dy[i];
44             if ((x == 0 || x == n + 1 || y == 0 || y == n + 1 || map[x][y] == '#') && vis[p.x][p.y][i] == 0 && vis[x][y][(i + 2) % 4] == 0) {
45                 ans++;
46                 vis[p.x][p.y][i] = 1;
47                 vis[x][y][(i + 2) % 4] = 1;
48             }
49             if (ok(x, y) && map[x][y] == '.' && num[x][y] == 0) {
50                 num[x][y] = 1;
51                 q.push(point(x, y));
52             }
53         }
54     }
55     return ans;
56 }
57
58 int main() {
59     while (scanf("%d\n", &n) != EOF) {
60         for (int i = 1; i <= n; i++)
61             gets(map[i] + 1);
62         for (int i = 0; i <= n + 2; i++)
63             for (int j = 0; j <= n + 2; j++) {
64                 num[i][j] = 0;
65                 for (int k = 0; k < 4; k++)
66                     vis[i][j][k] = 0;
67             }
68         int ans=bfs(point(1,1))+bfs(point(n,n));
69         printf("%d\n", ans*9-36);
70     }
71     return 0;
72 }

晓天_xiaotian 2010-09-29 11:23 发表评论

Timus 1016. A Cube on the Walk

晓天_xiaotian — Tue, 28 Sep 2010 16:07:00 GMT

摘要: bfs，很基础的。可以看作在一个图上跑最短路，关健是建图，本人愚拙，手打整个图。 Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/--> 1 /* 2 * File:&n... 阅读全文

晓天_xiaotian 2010-09-29 00:07 发表评论