BZOJ1486:[HNOI2009]最小圈(最短路,二分)

Description

Input

Output

Sample Input

4 5
1 2 5
2 3 5
3 1 5
2 4 3
4 1 3

Sample Output

3.66666667

Solution

二分一个$ans$，然后每条边减去$ans$，若存在负环则$ans$合法。

判断个入队20次就很稳了。

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #define N (3009)
 using namespace std;

 double eps=1e-,dis[N];
 struct Edge{int to,next,len;}edge[N*];
 int n,m,u,v,l,cnt[N],used[N],head[N],num_edge;
 queue<int>q;

 void add(int u,int v,int l)
 {
     edge[++num_edge].to=v;
     edge[num_edge].len=l;
     edge[num_edge].next=head[u];
     head[u]=num_edge;
 }

 bool SPFA(double mid)
 {
     memset(cnt,,sizeof(cnt));
     memset(used,,sizeof(used));
     for (int i=; i<=n; ++i) dis[i]=1e18;
     while (!q.empty()) q.pop();
     dis[]=; used[]=true; q.push();
     while (!q.empty())
     {
         int x=q.front(); q.pop();
         for (int i=head[x]; i; i=edge[i].next)
             if (dis[x]+edge[i].len-mid<dis[edge[i].to])
             {
                 dis[edge[i].to]=dis[x]+edge[i].len-mid;
                 if (!used[edge[i].to])
                 {
                     cnt[edge[i].to]++;
                     if (cnt[edge[i].to]>) return true;
                     used[edge[i].to]=true;
                     q.push(edge[i].to);
                 }
             }
         used[x]=false;
     }
     return false;
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     for (int i=; i<=m; ++i)
         scanf("%d%d%d",&u,&v,&l),add(u,v,l);
     double l=-1e7, r=1e7;
     while (r-l>eps)
     {
         double mid=(l+r)/;
         if (SPFA(mid)) r=mid;
         else l=mid;
     }
     printf("%.8lf\n",l);
 }