题目大意：给出一些作物，这些作物要不就是种在A地，要不就是种在B地，有些作物种在一起会有额外收成。问最多可以获得多少收成。

思路：最小割模型，与S集相连的点都是种在A地的点，与T集相连的点都是种在B地的点。中间随便乱搞一下，总之最后就是所有收成-最大流就是最后答案。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 3010
#define MAXE 5000010
#define S 0
#define T (MAX - 1)
#define INF 0x3f3f3f3f
using namespace std;
 
struct MaxFlow{
    int head[MAX],total;
    int next[MAXE],aim[MAXE],flow[MAXE];
     
    int deep[MAX];
     
    MaxFlow() {
        total = 1;
    }
    void Add(int x,int y,int f) {
        next[++total] = head[x];
        aim[total] = y;
        flow[total] = f;
        head[x] = total;
    }
    void Insert(int x,int y,int f) {
        Add(x,y,f);
        Add(y,x,0);
    }
    bool BFS() {
        static queue<int> q;
        while(!q.empty())   q.pop();
        memset(deep,0,sizeof(deep));
        deep[S] = 1;
        q.push(S);
        while(!q.empty()) {
            int x = q.front(); q.pop();
            for(int i = head[x]; i; i = next[i])
                if(flow[i] && !deep[aim[i]]) {
                    deep[aim[i]] = deep[x] + 1;
                    q.push(aim[i]);
                    if(aim[i] == T) return true;
                }
        }
        return false;
    }
    int Dinic(int x,int f) {
        if(x == T)  return f;
        int temp = f;
        for(int i = head[x]; i; i = next[i])
            if(flow[i] && temp && deep[aim[i]] == deep[x] + 1) {
                int away = Dinic(aim[i],min(temp,flow[i]));
                if(!away)   deep[aim[i]] = 0;
                flow[i] -= away;
                flow[i^1] += away;
                temp -= away;
            }
        return f - temp;
    }
}solver;
 
int cnt,pos;
int asks,ans;
 
int main()
{
    cin >> cnt;
    pos = cnt;
    for(int x,i = 1; i <= cnt; ++i) {
        scanf("%d",&x);
        ans += x;
        solver.Insert(S,i,x);
    }
    for(int x,i = 1; i <= cnt; ++i) {
        scanf("%d",&x);
        ans += x;
        solver.Insert(i,T,x);
    }
    cin >> asks;
    for(int num,x,i = 1; i <= asks; ++i) {
        scanf("%d",&num);
        pos += 2;
        scanf("%d",&x);
        solver.Insert(S,pos - 1,x);
        ans += x;
        scanf("%d",&x);
        solver.Insert(pos,T,x);
        ans += x;
        for(int j = 1; j <= num; ++j) {
            scanf("%d",&x);
            solver.Insert(pos - 1,x,INF);
            solver.Insert(x,pos,INF);
        }
    }
    int max_flow = 0;
    while(solver.BFS())
        max_flow += solver.Dinic(S,INF);
    cout << ans - max_flow << endl;
    return 0;
}

BZOJ 3438 小M的作物 最小割