POJ1734 Sightseeing trip floyd求最小环问题

问题描述

m个点， m条边， 求图中存在的路径最小的环

样例

Sample Input
5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20
Sample Output
1 3 5 2

思路
    用ans存最小环的距离，dis[i, j]存i 到 j 的最短路径， pass[i, j] 存 i 到 j 最短距离经过哪些点， path存最终的路径

代码

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>

using namespace std;

const int N = 110;

int g[N][N], dis[N][N], pass[N][N];
vector<int> path;
int n, m, ans = 0x3f3f3f3f;

void get_path(int x, int y)
{
    if (pass[x][y] == 0)
        return;
    get_path(x, pass[x][y]);
    path.push_back(pass[x][y]);
    get_path(pass[x][y], y);
}

int main()
{
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    for (int i = 0; i <= n; i++)
        g[i][i] = 0;

    for (int i = 1; i <= m; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = min(g[a][b], c);
    }
    memcpy(dis, g, sizeof g);
    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i < k; i++)
            for (int j = i + 1; j < k; j++)
            {
                if ((long long)dis[i][j] + g[i][k] + g[k][j] < ans)
                {
                    ans = dis[i][j] + g[i][k] + g[k][j];
                    path.clear();
                    path.push_back(i);
                    get_path(i, j);
                    path.push_back(j);
                    path.push_back(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];
                    pass[i][j] = k;
                }
    }

    if (ans == 0x3f3f3f3f)
    {
        puts("No solution.");
    }
    else
    {
        for (int i = 0; i < path.size(); i++)
            cout << path[i] << " ";
        cout << endl;
    }

    return 0;
}