[BZOJ1497][NOI2006]最大获利（最小割）

题目描述

传送门

题解

对于每一个中转站，s->i,pi，割掉表示花费pi建立中转站

对于每一个顾客，i->t,ci，割掉表示放弃ci的收益

如果一个顾客x需要某一个中转站y，那么y->x,inf，因为py和cx不能同时满足，必须选一条割掉

所有顾客的收益之和减去最小割即为答案

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
#define N 100005
#define inf 1000000000

int n,m,p,a,b,c,s,t,sum,maxflow;
int tot,point
,nxt[N*4],v[N*4],remain[N*4];
int deep
,cur
,last
,num
;
queue <int> q;

void addedge(int x,int y,int cap)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;
}
void bfs(int t)
{
for (int i=s;i<=t;++i) deep[i]=t;
deep[t]=0;
for (int i=s;i<=t;++i) cur[i]=point[i];
while (!q.empty()) q.pop();
q.push(t);

while (!q.empty())
{
int now=q.front();q.pop();
for (int i=point[now];i!=-1;i=nxt[i])
if (deep[v[i]]==t&&remain[i^1])
{
deep[v[i]]=deep[now]+1;
q.push(v[i]);
}
}
}
int addflow(int s,int t)
{
int now=t,ans=inf;
while (now!=s)
{
ans=min(ans,remain[last[now]]);
now=v[last[now]^1];
}
now=t;
while (now!=s)
{
remain[last[now]]-=ans;
remain[last[now]^1]+=ans;
now=v[last[now]^1];
}
return ans;
}
void isap(int s,int t)
{
bfs(t);
for (int i=s;i<=t;++i) ++num[deep[i]];
int now=s;

while (deep[s]<t)
{
if (now==t)
{
maxflow+=addflow(s,t);
now=s;
}

bool has_find=false;
for (int i=cur[now];i!=-1;i=nxt[i])
if (deep[v[i]]+1==deep[now]&&remain[i])
{
has_find=true;
cur[now]=i;
last[v[i]]=i;
now=v[i];
break;
}

if (!has_find)
{
int minn=t-1;
for (int i=point[now];i!=-1;i=nxt[i])
if (remain[i]) minn=min(minn,deep[v[i]]);
if (!(--num[deep[now]])) break;
++num[deep[now]=minn+1];
cur[now]=point[now];
if (now!=s) now=v[last[now]^1];
}
}
}

int main()
{
tot=-1;memset(point,-1,sizeof(point));
scanf("%d%d",&n,&m);
s=1,t=n+m+2;
for (int i=1;i<=n;++i)
{
scanf("%d",&p);
addedge(s,1+i,p);
}
for (int i=1;i<=m;++i)
{
scanf("%d%d%d",&a,&b,&c);
addedge(1+n+i,t,c);
addedge(1+a,1+n+i,inf);
addedge(1+b,1+n+i,inf);
sum+=c;
}
isap(s,t);
printf("%d\n",sum-maxflow);
}