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 = 1000 + 10;
int n, m, k, Case = 0, G[maxn][maxn], dist[maxn];
bool vis[maxn];

int Dijkstra(int source, int aid) {
    for(int i = 1; i <= n; i ++)
        dist[i] = (i == source ? INF : 0);
    for(int i = 1; i <= n; i ++) {
        int minf = -1;
        for(int j = 1; j <= n; j ++)
            if(!vis[j] && minf < dist[j]) {
                minf = dist[j];
                k = j;
            }
        vis[k] = true;
        if(minf == - 1) break;
        for(int j = 1; 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 = 1; i <= n; i ++) {
            for(int j = 1; j <= n; j ++)
                G[i][j] = 0;
        }
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= m; i ++) {
            scanf("%d %d %d", &x, &y, &w);
            G[x][y] = G[y][x] = max(G[x][y], w);
        }
        printf("Scenario #%d:
%d

", ++Case, Dijkstra(1, n));
    }
    return 0;
}