杭电Problem-3790 最短路径问题

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最短路径问题

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20086 Accepted Submission(s): 5971
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Problem Description

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

Input

输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点。n和m为0时输入结束。
(1<n<=1000, 0<m<100000, s != t)

Output

输出一行有两个数，最短距离及其花费。

Sample Input

3 21 2 5 62 3 4 51 30 0

Sample Output

9 11

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#define min(a, b) (a < b) ? a: b#define MAX_N 1005using namespace std;const int INF = 9999999;bool used[MAX_N];int cost[MAX_N][MAX_N];int d[MAX_N];int mincost[MAX_N];int dis[MAX_N][MAX_N];int main(){int n, m, a, b, cos, s, t, disrt;while (scanf("%d %d", &n, &m), n && m){for (int i = 0; i <= n; i++) {for (int j = 0; j <= n; j++) {if(i == j){dis[i][j] = dis[i][j] = 0;cost[i][j] = cost [j][i] = 0;}elsedis[j][i] = dis[i][j] = cost[i][j] = cost[j][i] = INF;}}for (int i = 1; i <= m; i++) {scanf("%d %d %d %d", &a, &b, &disrt, &cos);if(dis[a][b] > disrt) {dis[a][b] = dis[b][a] = disrt;cost[a][b] = cost[b][a] = cos;}else if (dis[a][b] == disrt) {cost[a][b] = cost[b][a] = min(cost[a][b], cos);}}scanf("%d %d", &s, &t);memset(d,INF,sizeof(d));memset(used,false,sizeof(used));memset(mincost,INF,sizeof(mincost));d[s] = 0;mincost[s] = 0;while (true) {int v = -1;for (int u = 1; u <= n; u++) {if(!used[u]&&(v == -1 || d[v] > d[u]))v = u;}if (v == -1)break;used[v] = true;for (int u = 1; u <= n; u++) {if (d[u] > d[v] + dis[u][v]) {d[u] = d[v] + dis[u][v];    mincost[u] = cost[v][u] + mincost [v];}else if (d[u] == d[v] + dis[u][v]) {mincost[u] = min(mincost[u], mincost[v] + cost[u][v]);}}}printf("%d %d\n", d[t], mincost[t]);}return 0;}

对于数据的初始话，我就呵呵了，一直WA了十多遍，看别人的代码才改过来。。。。。

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