Floyd求最小环

Floyd求最小环

原理:

1. 当第三层循环到点k时, 所有点之间的最短路只通过[1,k)更新, 记$dis_{u,v}$是从u到v且仅经过编号在区间$[1,k)$中的点的最短路。
2. 最小环至少要有3个顶点, 最小环通过一条最短路和枚举一个点的2条边来更新 $ans=min(ans,dis_{u,v}+g[u][k]+g[k][v])$

算法的疑问?

1. 如果g[u][k]+g[k][v]改为dis[u][k]+dis[k][v]错误: 因为u到v的最短路可能会重复，不成环,只能用边来枚举
2. 至少有三个点,所以三层循环i!=j

POJ 1734

• path记录路径
• pre[i][j]记录：最短路径中j点的上一个点是谁

#include <cstring>
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 110;
const int inf  = 9999999;
int n,m,tot;
int g[maxn][maxn],dis[maxn][maxn];
int pre[maxn][maxn],path[maxn];

int main() {
    cin >> n >> m;
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=n; ++j) {
            g[i][j] = dis[i][j] = inf;
            pre[i][j] = i;
        }
    for (int i=1; i<=m; ++i) {
        int u,v,w;
        cin >> u >> v >> w;
        g[u][v] = g[v][u]= dis[u][v] = dis[v][u] = min(g[u][v],w);
    }
    int ans = inf;
    for (int k=1; k<=n; ++k) {
        for (int i=1; i<k; ++i)
            for (int j=i+1;j<k; ++j) {
                int temp = dis[i][j] + g[i][k] + g[k][j];
                if (temp < ans) {
                    ans = temp;
                    tot = 0;
                    int p = j;
                    while (p != i) {
                        path[tot++] = p;
                        p = pre[i][p];
                    }
                    path[tot++] = i;
                    path[tot++] = k;
                }
            }

        for (int i=1; i<=n; ++i)
            for (int j=1; j<=n; ++j)
                if (dis[i][j] > dis[i][k] + dis[k][j]) {
                    dis[i][j] = dis[i][k] + dis[k][j];
                    pre[i][j] = pre[k][j];
                }
    }
    if(ans==inf)
        cout << "No solution." << endl;
    else {
        for (int i=0; i<tot; ++i)
            cout << path[i] << " ";
        cout << endl;
    }
    return 0;
}