題意：

　　給一個有向圖，問這個圖構成的所有的環中，平均權值最小是多少？

　　平均權值是，回路上權值和除以邊數。

分析：

　　二分答案，讓每條邊減去二分的值，然後用Bellman-Ford算法，判斷負環即可。

坑點：

　　如果判斷出隊次數，則要看是否大於n次，如果判斷入隊次數，則要看是否大於等於n次。

代碼：

  1 #include<iostream>
  2 using namespace std;
  3 #include<cstdio>
  4 #include<cmath>
  5 #include<vector>
  6
 
 #include<cstring>
  7 #include<queue>
  8 #define MAXN 500
  9 typedef long long LL;
 10 int ii=0;
 11 struct teamdata{
 12     double d;int point;
 13     bool operator<(const teamdata& pt)const{
 14         return this->d>pt.d;
 15     }
 16 };
 17 struct bian{
 18
 
     int point;
 19     double d;
 20 };
 21 int n,m,k,l,s,t;
 22 vector<bian> maps[MAXN];
 23 const double inf=1e10;
 24 double d[MAXN];
 25 int pre[MAXN];
 26 priority_queue<teamdata> team;
 27 int v[MAXN];
 28 int SPFA(double mid){
 29     LL ans=-1;
 30     teamdata pt;
 31
 
     while(!team.empty()) team.pop();
 32     teamdata pd;
 33     for (int i=0;i<=MAXN-1;i++) d[i]=inf,pre[i]=0,v[i]=0;d[s]=0;
 34     for (int i=1;i<=n;i++) {
 35         pt.d=0;pt.point=i;team.push(pt);d[i]=0;v[i]=1;
 36     }
 37     while(!team.empty()){
 38         pt=team.top();team.pop();
 39         //cout<<"*"<<pt.point<<" "<<pt.d<<endl;
 40         v[pt.point]=0;
 41         for (int i=0;i<maps[pt.point].size();i++){
 42             int nb=maps[pt.point][i].point;
 43             if (d[nb]>d[pt.point]+maps[pt.point][i].d-mid){
 44                 d[nb]=d[pt.point]+maps[pt.point][i].d-mid;
 45                 pd.d=d[nb];pd.point=nb;
 46                 if (v[nb]==0){
 47                     v[nb]=1;
 48                 //printf("\t%d %lf %lf\n",pd.point,pd.d,mid);
 49                     team.push(pd);
 50         pre[nb]++;
 51         if (pre[nb]>=n) {
 52        return 1;
 53         }
 54                 }
 55         }
 56         }
 57     }
 58     return 0;
 59 }
 60 void deal(){
 61     scanf("%d%d",&n,&m);
 62     double w;
 63     for (int i=0;i<=n;i++) maps[i].clear();
 64     double ll=0;
 65     for (int i=1;i<=m;i++){
 66         int l,r;
 67         scanf("%d%d%lf",&l,&r,&w);
 68         bian pt;pt.d=w;pt.point=r;
 69         ll=max(ll,w);
 70         maps[l].push_back(pt);
 71     }
 72     double l,r,mid,ans=1e9;
 73     l=0;r=ll+1;
 74     if (SPFA(ll+1)==0){
 75         printf("No cycle found.\n");
 76         return;
 77     }
 78     else
 79         while((r-l)>1e-10){
 80         mid=(l+r)/2;
 81         int zans=0;
 82         zans=SPFA(mid);
 83         if (zans){
 84             if (mid<ans) ans=mid;
 85             r=mid;
 86         }else{
 87             l=mid;
 88         }
 89         }
 90     printf("%.2lf\n",ans);
 91 }
 92 int main(){
 93     int t;
 94     scanf("%d",&t);
 95     while(t--){
 96         ii++;
 97         printf("Case #%d: ",ii);
 98         deal();
 99     }
100     return 0;
101 }

UVA - 11090 Going in Cycle!!