POJ 3621-Sightseeing Cows-最优比率环｜SPFA+二分

最优比率环问题。二分答案，对于每一个mid，把节点的happy值归类到边上。

对于每条边，用mid×weight减去happy值，如果不存在负环，说明还可以更大。

 /*--------------------------------------------------------------------------------------*/
 //        Helica's header
 //        Second Edition
 //        2015.11.7
 //
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <ctype.h>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 #include <string>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <set>
 #include <map>

 //debug function for a N*M array
 #define debug_map(N,M,G) printf("\n");for(int i=0;i<(N);i++)\
 {for(int j=;j<(M);j++){\
 printf("%d",G[i][j]);}printf("\n");}
 //debug function for int,float,double,etc.
 #define debug_var(X) cout<<#X"="<<X<<endl;
 /*--------------------------------------------------------------------------------------*/
 using namespace std;

 const int INF = 0x3f3f3f3f;
 const int maxn = 1e3+;
 const double eps = 1e-;
 int N,M;
 int happy[maxn];

 struct Edge
 {
     int to,next,w;
 }edge[*maxn];

 int head[maxn],cnt[maxn],tol;
 double dist[maxn];
 bool vis[maxn];

 bool SPFA(double mid)
 {
     memset(vis,false,sizeof(vis));
     memset(cnt,,sizeof(cnt));
     for(int i=;i<=N;i++)dist[i]=INF;
     dist[]=;
     queue<int>Q;
     Q.push();
     while(!Q.empty()){
         int u=Q.front();
         Q.pop();
         vis[u]=false;
         cnt[u]++;
         if(cnt[u]>N)return true;
         for(int i=head[u];i!=-;i=edge[i].next){
             int v=edge[i].to,w=edge[i].w;
             double tmp=mid*w-happy[v];
             if(dist[u]+tmp<dist[v]){
                 dist[v]=dist[u]+tmp;
                 if(!vis[v]){ vis[v]=true;Q.push(v); }
             }
         }
     }
     return false;
 }
 void add_edge(int u,int v,int w)
 {
     edge[tol].to = v;
     edge[tol].w = w;
     edge[tol].next = head[u];
     head[u] = tol++;
 }

 int main()
 {
     //freopen("input.in","r",stdin);
     while(~scanf("%d%d",&N,&M))
     {
         memset(head,-,sizeof head);
         tol = ;
         for(int i=;i<=N;i++)    scanf("%d",&happy[i]);
         for(int i=,u,v,w;i<M;i++)
         {
             scanf("%d%d%d",&u,&v,&w);
             add_edge(u,v,w);
         }
         double L = ,R = 10000.0,mid;

         while(abs(L-R)>eps)
         {
             mid = (L+R)/2.0;
             if(SPFA(mid)) L = mid;
             else          R = mid;
         }

         printf("%.2f\n",mid);
     }
 }