poj_1125_Stockbroker Grapevine

简单的folyd,题意为：给定一个有向图，要求从那个点开始，获得的最少时间，输入起始点和所需要的最少时间。#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define MAXN    101int map[MAXN][MAXN];void folyd(const int &n){        for(int k = 1; k <= n; k ++) {                for(int i = 1; i <= n; i ++) {                        for(int j = 1; j <= n; j ++) {                                map[i][j] = min(map[i][j], map[i][k]+map[k][j]);                        }                }        }}int cal(const int &u, const int &n){        int rst(0);        for(int v = 1; v <= n; v ++) {                if( v == u ) {                        continue;                }                rst = max(rst, map[u][v]);        }        return rst;}int main(int argc, char const *argv[]){#ifndef ONLINE_JUDGE        freopen("test.in", "r", stdin);#endif        int n, v, cost, m, start, ans, tmp;        while( scanf("%d", &n) && n ) {                memset(map, 0x3f, sizeof(map));                for(int u = 1; u <= n; u ++) { scanf("%d", &m);                        for(; m; m --) {                                scanf("%d %d", &v, &cost); map[u][v] = cost;                        }                }                folyd(n); ans = 0x3f3f3f3f;                for(int i = 1; i <= n; i ++) {                        if( ans > (tmp = cal(i, n)) ) {                                ans = tmp; start = i;                        }                }                printf("%d %d\n", start, ans);        }        return 0;}