POJ 1734 Sightseeing trip Floyd求最小环

题目大意：给一张无向图，求这个图中的最小环并输出路径。

思路：在每次正常的Floyd更新最短路之前，先判断是否有最小环，然后再进行正常的floyd更新。如果在k更新最短路之前i和j就已经有点将他们更新，那么i->j就有一个环。然后就可以递归找出路径。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 110
#define INF 0x2a2a2a2a
using namespace std;

int points,edges;
int map[MAX][MAX],src[MAX][MAX],cover_by[MAX][MAX];
int ans[MAX],total;
int min_circle = INF;

void Floyd();
void GetAns(int l,int r);

int main()
{
	cin >> points >> edges;
	memset(map,0x2a,sizeof(map));
	for(int x,y,z,i = 1;i <= edges; ++i) {
		scanf("%d%d%d",&x,&y,&z);
		map[x][y] = map[y][x] = min(map[x][y],z);
	}
	memcpy(src,map,sizeof(src));
	Floyd();
	if(min_circle == INF)	puts("No solution.");
	else
		for(int i = 1;i <= total; ++i) 
			printf("%d%c",ans[i],i == total ? '\n':' ');
	return 0;
}

void Floyd()
{
	for(int k = 1;k <= points; ++k) {
		for(int i = 1;i <= points; ++i)
			for(int j = 1;j <= points; ++j) {
				if(min_circle > src[i][k] + src[k][j] + map[i][j] && i != j && i != k && j != k) {
					min_circle = src[i][k] + src[k][j] + map[i][j];
					total = 0;
					ans[++total] = i;
					GetAns(i,j);
					ans[++total] = j;
					ans[++total] = k;
				}
			}
		for(int i = 1;i <= points; ++i)
			for(int j = 1;j <= points; ++j)
				if(map[i][j] > map[i][k] + map[k][j]) {
					map[i][j] = map[i][k] + map[k][j];
					cover_by[i][j] = cover_by[j][i] = k;
				}
	}
}

void GetAns(int l,int r)
{
	int mid = cover_by[l][r];
	if(!mid)	return ;
	GetAns(l,mid);
	ans[++total] = mid;
	GetAns(mid,r);
}