BZOJ3130: [Sdoi2013]费用流[最大流 实数二分]

//
//  main.cpp
//  sdoi2003费用流
//
//  Created by Candy on 25/11/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=105,M=1005,INF=1e9;
const double eps=1e-5;
int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();}
    return x*f;
}
int n,m,p,u,v,c,s,t;
struct data{
    int u,v,c;
}a[M];
struct edge{
    int v,ne;
    double c,f;
}e[M<<1];
int h[N],cnt=0;
inline void ins(int u,int v,double c){
    cnt++;
    e[cnt].v=v;e[cnt].c=c;e[cnt].f=0;e[cnt].ne=h[u];h[u]=cnt;
    cnt++;
    e[cnt].v=u;e[cnt].c=0;e[cnt].f=0;e[cnt].ne=h[v];h[v]=cnt;
}
void build(double mid){
    cnt=0;
    memset(h,0,sizeof(h));
    for(int i=1;i<=m;i++) ins(a[i].u,a[i].v,min((double)a[i].c,mid));
}
int cur[N];
int d[N],vis[N],q[N],head,tail;
bool bfs(){
    memset(vis,0,sizeof(vis));
    memset(d,0,sizeof(d));
    head=tail=1;
    d[s]=0;vis[s]=1;q[tail++]=s;
    while(head!=tail){
        int u=q[head++];
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v;
            if(!vis[v]&&e[i].c>e[i].f){
                d[v]=d[u]+1;vis[v]=1;
                q[tail++]=v;
                if(v==t) return 1;
            }
        }
    }
    return 0;
}

double dfs(int u,double a){
    if(u==t||a==0) return a;
    double flow=0,f;
    for(int &i=cur[u];i;i=e[i].ne){
        int v=e[i].v;
        if(d[v]==d[u]+1&&(f=dfs(v,min(a,e[i].c-e[i].f)))>0){
            flow+=f;
            e[i].f+=f;
            e[((i-1)^1)+1].f-=f;
            a-=f;
            if(a==0) break;
        }
    }
    return flow;
}
double dinic(){
    double flow=0;
    while(bfs()){
        for(int i=1;i<=n;i++) cur[i]=h[i];
        flow+=dfs(s,INF);
    }
    return flow;
}
int main(int argc, const char * argv[]) {
    n=read();m=read();p=read();s=1;t=n;
    double l=0,r=0;
    for(int i=1;i<=m;i++){
        a[i].u=read(),a[i].v=read(),a[i].c=read();
        ins(a[i].u,a[i].v,a[i].c);r=max(r,(double)a[i].c);
    }
    //r+=eps;
    double old=dinic();
    while(r-l>eps){
        double mid=(l+r)*0.5;//printf("%f %f\n",l,r);
        build(mid);
        double mx=dinic();
        if(fabs(mx-old)<eps) r=mid;
        else l=mid;
    }
    printf("%d\n%.4f",(int)old,l*p);
    return 0;
}