做vijos1046時想最小環想了半天，還是沒想出來，到網上一搜原來是用floyd來求……

最小環

環，，i,j,k三點，環長則為三點 兩兩間距，

且路不能有交點，

所以，我們想到map[i][j] + dist[i][k] + dist[j][k]

map記錄最短路，dist記錄直接相連的邊

只要map[i][j]不過k點就算得出一個環

讓最短路不過k點

想到floyd的dp方程

結果呼之欲出了吧~~我就不說了，直接上虛擬碼

/*由於是無向圖所以計算可以少一半。。*/
dist = map;
k = 1->n;
{
i = 1->k-1;
j = i+1->k;
ans = min(ans,map[i][j] +dist[i][k] + dist[j][k] );//需注意的是i,j,k三點不能相同，所以迴圈終止是k-1,i-1。。。
i = 1->n;
j = 1->i;
dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]);
}
 

vijos1046程式碼，最小環實用

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
#define INF 10000000
#define MAX 101
int n,m,map[MAX][MAX];
int floyd()
{
	int mincircle = INF;
	int Dist[n+1][n+1];
	for (int i = 1; i <= n; i++)
	for (int j = 1; j <= n; j++)
	{
		Dist[i][j] = map[i][j];
	}
	for(int k = 1; k <= n; k++)
	{
		for(int i = 1; i < k; i++)
		for(int j = i+1; j < k; j++)
		{
			mincircle = min(mincircle,Dist[i][j]+map[j][k]+map[k][i]);
		}
		for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
		{
			Dist[i][j] = min(Dist[i][j],Dist[i][k] + Dist[k][j]);
		}
	}
	return mincircle;
}
void init()
{
	int b,e,l;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			map[i][j] = INF;
		}
		map[i][i] = 0;
	}
	for (int i = 1; i <= m; i++)
	{
		scanf("%d %d %d",&b,&e,&l);
		map[b][e] = map[e][b] = min(map[b][e],l);
	}
	int ans = floyd();
	if ( ans >= INF )
	{
		printf("No solution.");
	}
	else
	{
		printf("%d",ans);
	}
}
int main()
{
    freopen("in.txt","r",stdin);
	while(scanf("%d %d",&n,&m) != EOF)
	{
		init();
		printf("\n");
	}
	return 0;
}