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  • bzoj 3130: [Sdoi2013]费用流

      1 #include<cstdio>
      2 #include<iostream>
      3 #define M 10000
      4 #define inf 0x7fffffff
      5 #include<cstring>
      6 #define eps 1e-5
      7 using namespace std;
      8 struct data 
      9 {
     10     int x,y;
     11     double z;
     12 }a[M];
     13 int d[M],q[M],S,T,cnt=1,n,m,head[M],next[M],u[M];
     14 double p,w[M],l,r,ss,ans,ans1;
     15 void jia1(int a1,int a2,double a3)
     16 {
     17     cnt++;
     18     next[cnt]=head[a1];
     19     head[a1]=cnt;
     20     u[cnt]=a2;
     21     w[cnt]=a3;
     22     return;
     23 }
     24 void jia(int a1,int a2,double a3)
     25 {
     26     jia1(a1,a2,a3);
     27     jia1(a2,a1,0);
     28     return;
     29 }
     30 bool bfs()
     31 {
     32     memset(d,0,sizeof(int)*(T+1));
     33     int h=0,t=1;
     34     q[1]=S;
     35     d[S]=1;
     36     for(;h<t;)
     37       {
     38         h++;
     39         int p=q[h];
     40         for(int i=head[p];i;i=next[i])
     41           if(!d[u[i]]&&w[i])
     42             {
     43                 d[u[i]]=d[p]+1;
     44                 if(d[T])
     45                   return 1;
     46                 t++;
     47                 q[t]=u[i];
     48             }
     49       }
     50     return 0;
     51 }
     52 double dinic(int s,double f)
     53 {
     54     if(s==T)
     55       return f;
     56     double rest=f;
     57     for(int i=head[s];i&&rest;i=next[i])
     58       if(w[i]&&d[u[i]]==d[s]+1)
     59         {
     60             double now=dinic(u[i],min(rest,w[i]));
     61             if(!now)
     62               d[u[i]]=0;
     63             w[i]-=now;
     64             w[i^1]+=now;
     65             rest-=now;
     66         }
     67     return f-rest;  
     68 }
     69 void jian(double mi)
     70 {
     71     cnt=1;
     72     memset(head,0,sizeof(int)*(T+1));
     73     for(int i=1;i<=m;i++)
     74       if(a[i].z<mi)
     75         jia(a[i].x,a[i].y,a[i].z);
     76       else
     77         jia(a[i].x,a[i].y,mi);
     78     return;
     79 }
     80 int main()
     81 {
     82     scanf("%d%d%lf",&n,&m,&p);
     83     for(int i=1;i<=m;i++)
     84       {
     85         scanf("%d%d%lf",&a[i].x,&a[i].y,&a[i].z);
     86         jia(a[i].x,a[i].y,a[i].z);
     87       }
     88     S=1;
     89     T=n;
     90     for(;bfs();)
     91       ans+=dinic(S,inf);
     92     printf("%d
    ",(int)ans);
     93     l=0;
     94     r=5000000;
     95     for(;r-l>eps;)
     96       {
     97         double mid=(l+r)/2;
     98         jian(mid);
     99         ans1=0;
    100         for(;bfs();)
    101           ans1+=dinic(S,inf);
    102         if(ans1==ans)
    103           {
    104             ss=mid;
    105             r=mid;
    106           }
    107         else
    108           l=mid;
    109       }
    110     printf("%.4lf",ss*p);
    111     return 0;
    112 }

    贪心 BOB肯定全加在最大权值的边上,二分权值网络流。

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  • 原文地址:https://www.cnblogs.com/xydddd/p/5309073.html
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