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\n",(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。

posted on 2016-03-22 23:16  xiyuedong  阅读(137)  评论(0编辑  收藏  举报