POJ 1734.Sightseeing trip （Floyd 最小环）-优快云博客

本文详细介绍了Floyd算法解决最小环问题的过程，包括代码实现、时间复杂度分析及应用实例，帮助读者理解并掌握该算法。

Floyd 最小环模板题

code

/*
       floyd最小环，记录路径，时间复杂度O（n^3）
       不能处理负环
*/
#include <iostream>
#include <cstring>
using namespace std;
const int INF = 109, maxn = 252645135;
int g[INF][INF], dis[INF][INF], pre[INF][INF];
int ans[INF];
//pr[i][j]记录i到j最短路径的第一个点
int n, m, tem , tol;
int main() {
	while (cin >> n >> m) {
		tem = maxn;
		memset (g, 0xf, sizeof g);
		memset (dis, 0xf, sizeof dis);
		memset (pre, 0, sizeof pre);
		int x, y, z;
		for (int i = 1; i <= m; i++) {
			cin >> x >> y >> z;
			if (z < g[x][y])
				dis[x][y] = dis[y][x] = g[x][y] = g[y][x] = z;
			pre[x][y] = y, pre[y][x] = x;
		}
		for (int k = 1; k <= n; k++) {
			for (int i = 1; i < k; i++)
				for (int j = i + 1; j < k; j++) {
					if (tem > dis[i][j] + g[i][k] + g[k][j]) {
						tem = dis[i][j] + g[i][k] + g[k][j];
						int t = i;
						tol = 0;
						while (t != 0) {
							ans[++tol] = t;
							t = pre[t][j];
						}
						ans[++tol] = k;
					}
				}
			for (int i = 1; i <= n; i++)
				for (int j = 1; j <= n; j++)
					if (i != j && dis[i][k] + dis[k][j] < dis[i][j]) {
						dis[i][j] = dis[i][k] + dis[k][j];
						pre[i][j] = pre[i][k];
					}
		}
		if (tem == maxn) cout << "No solution.";
		else
			for (int i = 1; i <= tol; i++)
				cout << ans[i] << ' ';
		cout << endl;
	}
	return 0;
}

转载于:https://www.cnblogs.com/keam37/p/3952349.html