HD1599 find the mincost route(floyd + 最小环)

题目链接

题意：求最小环

第一反应时floyd判断，但是涉及到最少3个点，然后就不会了，又想的是 双联通分量，这个不知道为什么不对。

Floyd 判断 最小环

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 const int Max =  + ;
 int g[Max][Max], dist[Max][Max];
 //dist【i】【j】保存i到j的最短路经，然后i -> j -> k 就可以枚举k， dist[i][j] + g[i][k] + g[j][k]就是一个环的权值
 int mincost;
 void Floyed(int n)
 {
     mincost = INF;
     for (int k = ; k <= n; k++)
     {
         for (int i = ; i < k; i++)
         {
             for (int j = i + ; j < k; j++)
             {
                 if (dist[i][j] != INF && g[i][k] != INF && g[k][j] != INF)
                 {
                     int temp = dist[i][j] + g[i][k] + g[k][j]; // 原先这里直接相加，一直没找到错误，爆精度
                     if (temp < mincost)
                         mincost = temp;
                 }
             }
         }
         for (int i = ; i <= n; i++)
         {
             for (int j = ; j <= n; j++)
             {
                 if (dist[i][k] != INF && dist[k][j] != INF)
                     dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
             }
         }
     }
 }
 int main()
 {
     int n, m;
     while (scanf("%d%d", &n, &m) != EOF)
     {
         for (int i = ; i < Max; i++)
             for (int j = ; j < Max; j++)
             {
                 g[i][j] = INF;
                 dist[i][j] = INF;
             }
         int a, b, c;
         for (int i = ; i <= m; i++)
         {
             scanf("%d%d%d", &a, &b, &c);
             if (g[a][b] > c)
                 g[a][b] = g[b][a] = dist[a][b] = dist[b][a] = c;
         }
         Floyed(n);
         if (mincost >= INF)
             printf("It's impossible.\n");
         else
             printf("%d\n", mincost);

     }
     return ;
 }