bellman_ford寻找平均权值最小的回路

给定一个有向图，如果存在平均值最小的回路，输出平均值。

使用二分法求解，对于一个猜测值mid，判断是否存在平均值小于mid的回路

如果存在平均值小于mid的包含k条边的回路，那么有w1+w2+w3+...+wk < k * mid,即(w1-mid)+(w2-mid)+..(wk-mid)<0,

即判断是否存在负权回路即可。

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=2031&mosmsg=Submission+received+with+ID+14572787

 #include <stdio.h>
 #include <string.h>
 const int N = +;
 const int INF = <<;
 struct Edge
 {
     int u,v;
     double weight;
 }g[];
 double dist[N];
 void relax(int u, int v,double weight)
 {
     if(dist[v] > dist[u] + weight)
         dist[v] = dist[u] + weight;
 }
 bool bellman_ford(int n, int m)
 {
     int i,j;
     for(i=; i<n-; ++i)//n-1循环
         for(j=; j<m; ++j)//枚举所有的边去松弛最短路径
         {
             relax(g[j].u,g[j].v,g[j].weight);
         }
     bool flag = false;
     for(i=; i<m; ++i)
         if(dist[g[i].v] > dist[g[i].u] + g[i].weight)
         {
             flag = true;
             break;
         }
     return flag;
 }
 bool test(double x,int n, int m)
 {
     int i;
     for(i=; i<m; ++i)
         g[i].weight -= x;
     bool ret =  bellman_ford(n,m);
     for(i=; i<m; ++i)
         g[i].weight += x;
     return ret;
 }
 int main()
 {
     int n,m,i,t,tCase=;
     double l,r;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         for(i=; i<=n; ++i)
             dist[i] = INF;
         l = r = ;
         for(i=; i<m; ++i)
         {
             scanf("%d%d%lf",&g[i].u,&g[i].v,&g[i].weight);
             r = r > g[i].weight ? r : g[i].weight;
         }
         if(!test(r+,n,m))printf("Case #%d: No cycle found.\n",tCase++);
         else
         {
             double mid;

             while(r-l>0.001)//因为题目要求保留2位小数，所以当r-l>0.001时，l就是答案。
             {
                 double mid = (r + l ) / ;
                 if(test(mid,n,m))
                     r = mid;
                 else
                     l = mid;
             }
             printf("Case #%d: %.2lf\n",tCase++,l);
         }

     }
     return ;
 }