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 = 100 + 10;
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 = 1; k <= n; k++)
    {
        for (int i = 1; i < k; i++)
        {
            for (int j = i + 1; 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 = 1; i <= n; i++)
        {
            for (int j = 1; 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 = 1; i < Max; i++)
            for (int j = 1; j < Max; j++)
            {
                g[i][j] = INF;
                dist[i][j] = INF;
            }
        int a, b, c;
        for (int i = 1; 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.
");
        else
            printf("%d
", mincost);

    }
    return 0;
}