BZOJ 3438 小M的作物 最小割

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 0x3f3f3f3fusing 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;}