【二分答案】【最大流】bzoj3130 [Sdoi2013]费用流
二分最大的边的cap,记作Lim。
把所有的边的cap设为min(Lim,cap[i])。
Bob一定会把单位费用加到最大边上。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define EPS 0.000001
#define N 101
#define INF 2147483647.0
#define M 1001
typedef double db;
int n,m,K,S,T;
int v[M<<1],next[M<<1],first[N],en;
db cap[M<<1];
void AddEdge(int U,int V,db Cap)
{
v[en]=V; cap[en]=Cap; next[en]=first[U]; first[U]=en++;
v[en]=U; cap[en]=0; next[en]=first[V]; first[V]=en++;
}
queue<int>q;
int d[N],cur[N];
bool bfs()
{
memset(d,-1,sizeof(int)*(n+1));
d[S]=0; q.push(S);
while(!q.empty())
{
int U=q.front(); q.pop();
for(int i=first[U];i!=-1;i=next[i])
if(d[v[i]]==-1&&cap[i]>EPS)
{
d[v[i]]=d[U]+1;
q.push(v[i]);
}
}
return d[T]!=-1;
}
db dfs(int U,db a)
{
if(U==T||a<=EPS) return a;
db Flow=0.0,f;
for(int &i=cur[U];i!=-1;i=next[i])
if(d[v[i]]==d[U]+1&&(f=(dfs(v[i],min(a,cap[i]))))>EPS)
{
cap[i]-=f;
cap[i^1]+=f;
Flow+=f;
a-=f;
if(a<=EPS)
break;
}
if(Flow<=EPS) d[U]=-1;
return Flow;
}
db MaxFlow()
{
db Flow=0.0,tmp;
while(bfs())
{
memcpy(cur,first,sizeof(int)*(n+1));
while((tmp=dfs(S,INF))>EPS) Flow+=tmp;
}
return Flow;
}
int xs[M],ys[M],zs[M];
int ChuShi;
bool check(db Lim)
{
memset(first,-1,sizeof(int)*(n+1));
en=0;
for(int i=1;i<=m;++i)
AddEdge(xs[i],ys[i],min((db)zs[i],Lim));
db t=MaxFlow();
return fabs(t-(db)ChuShi)<=EPS?1:0;
}
int main()
{
// freopen("bzoj3130.in","r",stdin);
db r=0.0,l=0.0;
scanf("%d%d%d",&n,&m,&K);
S=1; T=n;
memset(first,-1,sizeof(int)*(n+1));
for(int i=1;i<=m;++i)
{
scanf("%d%d%d",&xs[i],&ys[i],&zs[i]);
AddEdge(xs[i],ys[i],(db)zs[i]);
r=max(r,(db)zs[i]);
}
ChuShi=(int)MaxFlow();
while(r-l>EPS)
{
db mid=(l+r)/2.0;
if(check(mid)) r=mid-EPS;
else l=mid+EPS;
}
printf("%d\n%.4lf\n",ChuShi,l*(db)K);
return 0;
}
——The Solution By AutSky_JadeK From UESTC
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