poj 1125 求最短路径

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题目大意：给出一组顶点和各顶点到达其他的直接距离，求每个点到其他点的最短距离中的最大值，然后在这些点里面最大值里面的最小值，而且要注意这个点要是能到达其他每个点。

解题思路：floyd算法求出每对点之间的最小值，然后枚举每个顶点，就这个顶点到其他点的距离中的最大值，而且要保证这个顶点都能到达其他每个点，求这些最大值里面的最小值和相应顶点。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn = 101;int graph[maxn][maxn], dp[maxn][maxn];int n, m;int main(){while(true){scanf("%d", &n);if(n == 0)break;memset(graph, -1, sizeof(graph));for(int i = 1; i <= n; i++){graph[i][i] = 0;scanf("%d", &m);for(int j = 0; j < m; j++){int index, val;scanf("%d %d", &index, &val);graph[i][index] = val;}}for(int k = 1; k <= n; k++){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(graph[i][k] != -1 && graph[k][j] != -1){if(graph[i][j] == -1)graph[i][j] = graph[i][k] + graph[k][j];elsegraph[i][j] = min(graph[i][j], graph[i][k] + graph[k][j]);}}}}int index = -1, pre = -1, tmp = -1;bool flag = true;for(int i = 1; i <= n; i++){flag = true;tmp = -1; for(int j = 1; j <= n; j++){if(i == j)continue;if(graph[i][j] == -1){flag = false;break;}tmp = max(tmp, graph[i][j]);}if(!flag)continue;if(pre == -1){pre = tmp;index = i;}else{if(pre > tmp){pre = tmp;index = i;}}}if(index == -1){printf("disjoint\n");continue;}printf("%d %d\n", index, pre);}return 0;}