[BZOJ1497][NOI2006]最大获利（最小割）

来源：互联网发布：温度数据采集程序编辑：程序博客网时间：2024/05/16 08:23

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 1000000000int n,m,p,a,b,c,s,t,sum,maxflow;int tot,point[N],nxt[N*4],v[N*4],remain[N*4];int deep[N],cur[N],last[N],num[N];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);}

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