[洛谷P1361]小M的作物

题目大意：将作物种在A，B两地，对于每种作物，种A，B分别有不同的收益，对于一些特殊的作物集合，共同种到A，B集合分别有一些额外收益。求最大收益。

题解：最小割，S向i连容量为$a_i$的边，i向T连容量为$b_i$，对于每个集合，建两个大点，把每组内的点与其连接，容量为inf，两个大点分别与原点或汇点相连，容量为$c_{i1}$和$c_{i2}$

卡点：1.把inf的边设成了双向边

C++ Code：

#include <cstdio>
#include <cstring>
#define N 4010
#define M 2003010
using namespace std;
const int inf = 0x3f3f3f3f;
int st, ed , sum;
int d[N], q[N], h, t;
int n, m, k, o, c1, c2;
int cnt = 2, head[N];
struct Edge {
	int to, nxt, w;
}e[M << 1];

inline int min(int a, int b) {return a < b ? a : b;}
void addE(int a, int b, int c){
	e[cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
	e[cnt ^ 1] = (Edge) {a, head[b], 0}; head[b] = cnt ^ 1;
	cnt += 2;
}
void addE_(int a, int b, int c){
	e[cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
	e[cnt ^ 1] = (Edge) {a, head[b], c}; head[b] = cnt ^ 1;
	cnt += 2;
}
bool bfs() {
	memset(d, 0, sizeof d);
	d[q[h = t = 0] = st] = 1;
	while (h <= t) {
		int u = q[h++];
		if (u == ed) return true;
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if ((!d[v]) && (e[i].w)){
				d[v] = d[u] + 1;
				q[++t] = v;
			}
		}
	}
	return d[ed];
}
int dfs(int u, int low) {
	if ((u == ed) || (low <= 0)) return low;
	int res = 0;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if ((d[v] == d[u] + 1) && (e[i].w)){
			int w = dfs(v, min(e[i].w, low - res));
			e[i].w -= w;
			e[i ^ 1].w += w;
			res += w;
			if (res == low) return res;
		}
	}
	if (!res) d[u] = -1;
	return res;
}
void dinic() {
	int ans = 0;
	while (bfs()) {
		ans += dfs(st, inf);
//		printf("%d
", ans);
	}
	printf("%d
", sum - ans);
}
int main() {
	scanf("%d", &n);
	st = 0; ed = 4009;
	for (int i = 1; i <= n; i++) {scanf("%d", &o); addE(st, i, o); sum += o;}
	for (int i = 1; i <= n; i++) {scanf("%d", &o); addE(i, ed, o); sum += o;}
	scanf("%d", &m);
	for (int l = 1; l <= m; l++) {
		scanf("%d%d%d", &k, &c1, &c2); sum += c1 + c2;
		addE(st, l + n, c1);
		addE(l + n + m, ed, c2);
		for (int i = 1; i <= k; i++) {scanf("%d", &o); addE(l + n, o, inf); addE(o, l + n + m, inf);}
	}
	dinic();
	return 0;
}