BZOJ 3130 [Sdoi2013]费用流 ——网络流
【题目分析】
很容易想到,可以把P放在流量最大的边上的时候最优。
所以二分网络流,判断什么时候可以达到最大流。
流量不一定是整数,所以需要实数二分,整数是会WA的。
【代码】
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
//#include <map>
#include <set>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1005
#define me 50005
#define inf 0x3f3f3f3f
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define eps 1e-8
void Finout()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#endif
}
int Getint()
{
int x=0,f=1; char ch=getchar();
while (ch<'0'||ch>'9') {if (ch=='-') f=-1; ch=getchar();}
while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int h[me],to[me<<1],ne[me<<1];
double fl[me<<1];
int en=0,S=0,T=me-1;
int lim[me];
void add(int a,int b,double c)
{
// cout<<a<<" "<<b<<" "<<c<<endl;
to[en]=b; ne[en]=h[a]; fl[en]=c; h[a]=en++;
to[en]=a; ne[en]=h[b]; fl[en]=0; h[b]=en++;
}
int map[maxn],tag=0;
bool tell()
{
queue <int> q;
memset(map,-1,sizeof map);
map[S]=0;
q.push(S);
while (!q.empty())
{
int x=q.front(); q.pop();
for (int i=h[x];i>=0;i=ne[i])
{
if (map[to[i]]==-1&&fl[i]>eps)
{
map[to[i]]=map[x]+1;
q.push(to[i]);
}
}
}
if (map[T]!=-1) return true;
return false;
}
double zeng(int k,double r)
{
if (k==T) return r;
double ret=0;
for (int i=h[k];i>=0&&ret<r;i=ne[i])
if (map[to[i]]==map[k]+1&&fl[i]>eps)
{
double tmp=zeng(to[i],min(fl[i],(double)r-ret));
ret+=tmp; fl[i]-=tmp; fl[i^1]+=tmp;
}
if (!ret) map[k]=-1;
return ret;
}
int n,a[maxn],b[maxn],c[maxn],m,p;
int main()
{
Finout();
memset(h,-1,sizeof h);
n=Getint(); m=Getint(); p=Getint();
S=1;T=n;
int maxx=0;
F(i,1,m)
{
a[i]=Getint();
b[i]=Getint();
c[i]=Getint();
add(a[i],b[i],(double)c[i]);
maxx=max(c[i],maxx);
}
double ans=0,tmp;
while (tell()) while(tmp=zeng(S,(double)inf)) ans+=tmp;
cout<<ans+eps<<endl;
double l=0,r=(double)maxx;
while (l+eps<r)
{
// cout<<l<<" "<<r<<endl;
double mid=(l+r)/2;
en=0;memset(h,-1,sizeof h);
F(i,1,m) add(a[i],b[i],min(mid,(double)c[i]));
double now=0,tmp;
while (tell()) while (tmp=zeng(S,(double)inf)) now+=tmp;
if (fabs(now-ans)<=eps) r=mid;
else l=mid;
}
printf("%.5f\n",l*p);
}

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