POJ-1797.HeavyTransportation(最长路中的最小权值)

　　本题思路：最短路变形，改变松弛方式即可，dist存的是源结点到当前结点的最长路的最小权值。

　　参考代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #define INF 0x3f3f3f3f
 using namespace std;
 
 const int maxn =  + ;
 int n, m, k, Case = , G[maxn][maxn], dist[maxn];
 bool vis[maxn];
 
 int Dijkstra(int source, int aid) {
     for(int i = ; i <= n; i ++)
         dist[i] = (i == source ? INF : );
     for(int i = ; i <= n; i ++) {
         int minf = -;
         for(int j = ; j <= n; j ++)
             if(!vis[j] && minf < dist[j]) {
                 minf = dist[j];
                 k = j;
             }
         vis[k] = true;
         if(minf == - ) break;
         for(int j = ; j <= n; j ++)
             if(!vis[j] && dist[j] < min(dist[k], G[k][j]))
                 dist[j] = min(dist[k], G[k][j]);
     }
     return dist[aid];
 }
 
 int main () {
     int t, x, y, w;
     scanf("%d", &t);
     while(t --) {
         memset(vis, false, sizeof vis);
         for(int i = ; i <= n; i ++) {
             for(int j = ; j <= n; j ++)
                 G[i][j] = ;
         }
         scanf("%d %d", &n, &m);
         for(int i = ; i <= m; i ++) {
             scanf("%d %d %d", &x, &y, &w);
             G[x][y] = G[y][x] = max(G[x][y], w);
         }
         printf("Scenario #%d:\n%d\n\n", ++Case, Dijkstra(, n));
     }
     return ;
 }