BZOJ 1497 [NOI2006]最大获利

传送门

题解：最大权闭合子图。将正向收益与S连，花费与T连。建立用户到他后继中转站容量为INF的边，保证不会被割。最后正向收益的和减去最小割就是答案。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <cstring>
#include <iomanip>
#include <set>
#include<ctime>
//CLOCKS_PER_SEC
#define se second
#define fi first
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Pii pair<int,int>
#define Pli pair<ll,int>
#define ull unsigned long long
#define pb push_back
#define fio ios::sync_with_stdio(false);cin.tie(0)
const double Pi=3.14159265;
const int N=4e6+10;
const ull base=163;
const int INF=0x3f3f3f3f;
using namespace std;
int head[100000],to[N],tot=0,nx[N],cap[N];
int s,t;
inline int read(){
    int x=0;char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
void add(int u,int v,int c){
    to[tot]=v;
    nx[tot]=head[u];
    cap[tot]=c;
    head[u]=tot++;
     
    to[tot]=u;
    nx[tot]=head[v];
    cap[tot]=0;
    head[v]=tot++;
}
int d[100000];
int cur[100000];
int bfs(){
    memset(d,-1,sizeof(d));
    queue<int>q;
    q.push(s);
    d[s]=1;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];~i;i=nx[i]){
            int v=to[i];
            if(d[v]==-1&&cap[i]>0){
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
    return d[t]!=-1;
}
int dfs(int s,int a){
    if(s==t||a==0)return a;
    int flow=0,f;
    for(int &i=cur[s];~i;i=nx[i]){
        int v=to[i];
        if(d[s]+1==d[v] && cap[i]>0 && (f=dfs(v,min(a,cap[i])))>0){
            flow+=f;
            cap[i]-=f;
            cap[i^1]+=f;
            a-=f;
            if(a==0)break;
        }
    }
    return flow;
}
int dinic(){
    int ans=0;
    while(bfs()){
        for(int i=0;i<=t;i++)cur[i]=head[i];
        while(int di=dfs(s,INF)){
            ans+=di;
        }
    }
    return ans;
}
int p[N];
int main(){
    int n,m;
    n=read(),m=read();
    memset(head,-1,sizeof(head));
    s=0,t=60000;
    for(int i=1;i<=n;i++){
        p[i]=read();
        add(i,t,p[i]);
    }
    int o=n+1;
    int sum=0;
    while(m--){
        int l,r,c;
        l=read(),r=read(),c=read();
        sum+=c;
        add(s,o,c);
        add(o,l,INF);
        add(o,r,INF);
        o++;
    }
    cout<<sum-dinic();
    return 0;
}