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

Posted 2021-02-16 bianjunting

　　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：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define INF 0x3f3f3f3f
 5 using namespace std;
 6 
 7 const int maxn = 1000 + 10;
 8 int n, m, k, Case = 0, G[maxn][maxn], dist[maxn];
 9 bool vis[maxn];
10 
11 int Dijkstra(int source, int aid) {
12     for(int i = 1; i <= n; i ++)
13         dist[i] = (i == source ? INF : 0);
14     for(int i = 1; i <= n; i ++) {
15         int minf = -1;
16         for(int j = 1; j <= n; j ++)
17             if(!vis[j] && minf < dist[j]) {
18                 minf = dist[j];
19                 k = j;
20             }
21         vis[k] = true;
22         if(minf == - 1) break;
23         for(int j = 1; j <= n; j ++)
24             if(!vis[j] && dist[j] < min(dist[k], G[k][j]))
25                 dist[j] = min(dist[k], G[k][j]);
26     }
27     return dist[aid];
28 }
29 
30 int main () {
31     int t, x, y, w;
32     scanf("%d", &t);
33     while(t --) {
34         memset(vis, false, sizeof vis);
35         for(int i = 1; i <= n; i ++) {
36             for(int j = 1; j <= n; j ++)
37                 G[i][j] = 0;
38         }
39         scanf("%d %d", &n, &m);
40         for(int i = 1; i <= m; i ++) {
41             scanf("%d %d %d", &x, &y, &w);
42             G[x][y] = G[y][x] = max(G[x][y], w);
43         }
44         printf("Scenario #%d:
%d

", ++Case, Dijkstra(1, n));
45     }
46     return 0;
47 }

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