【图论】最小环算法

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做vijos1046时想最小环想了半天，还是没想出来，到网上一搜原来是用floyd来求……

转自http://blog.csdn.net/loriex/article/details/19171415

最小环

环，，i,j,k三点，环长则为三点两两间距，

且路不能有交点，

所以，我们想到map[i][j] + dist[i][k] + dist[j][k]

map记录最短路，dist记录直接相连的边

只要map[i][j]不过k点就算得出一个环

让最短路不过k点

想到floyd的dp方程

结果呼之欲出了吧~~我就不说了，直接上伪代码

/*由于是无向图所以计算可以少一半。。*/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三点不能相同，所以循环终止是k-1,i-1。。。i = 1->n;j = 1->i;dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]);}

vijos1046代码，最小环实用

#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;#define INF 10000000#define MAX 101int 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;}

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