## 题目链接：

http://exam.upc.edu.cn/problem.php?id=5120

## 题目：

### 题目描述

Open-pit mining is a surface mining technique of extracting rock or minerals from the earth by their removal from an open pit or borrow. Open-pit mines are used when deposits of commercially useful minerals or rocks are found near the surface. Automatic Computer Mining (ACM) is a company that would like to maximize
its profits by open-pit mining. ACM has hired you to write a program that will determine the maximum profit it can achieve given the description of a piece of land.
Each piece of land is modelled as a set of blocks of material. Block i has an associated value (vi), as well as a cost (ci), to dig that block from the land. Some blocks obstruct or bury other blocks. So for example if block i is obstructed by blocks j and k, then one must first dig up blocks j and k before block i can be dug up. A block can be dug up when it has no other blocks obstructing it.

### 输入

The first line of input is an integer N (1≤N≤200) which is the number of blocks. These blocks are numbered 1 through N.
Then follow N lines describing these blocks. The ith such line describes block i and starts with two integers vi, ci denoting the value and cost of the ith block (0≤vi,ci≤200).
Then a third integer 0≤mi≤N-1 on this line describes the number of blocks that block i obstructs.
Following that are mi distinct space separated integers between 1 and N (but excluding i) denoting the label(s) of the blocks that block i obstructs.
You may assume that it is possible to dig up every block for some digging order. The sum of values mi over all blocks i will be at most 500.

### 输出

Output a single integer giving the maximum profit that ACM can achieve from the given piece of land.

### 样例输入

5
0 3 2 2 3
1 3 2 4 5
4 8 1 4
5 3 0
9 2 0

### 样例输出

2

## 题意：

有n种矿石，第i种矿石的价值为v[i]，开采的花费为c[i]，这个矿石的下方压着m[i]种矿石，随后是m[i]个整数，代表压着的矿石的编号。如果矿石a压着矿石b，那么得先开采矿石a才能再开采矿石b，问能获得的最大权值为多少。

最大权闭合子图的裸题。

## 实现：

#include <bits/stdc++.h>
const int maxn = 507, maxm = 505 * 505 * 2;
struct { int next, to, flow; } edge[maxm << 1];
void addedge(int u, int v, int c) {
}
int dist[maxn], S, T, n, sum;
bool bfs(int S, int T) {
memset(dist, -1, sizeof(dist));
std::queue<int> que;
dist[S] = 0;
que.push(S);
int u, v;
while (!que.empty()) {
u = que.front(), que.pop();
for (int i = head_edge[u]; ~i; i = edge[i].next) {
v = edge[i].to;
if (dist[v] == -1 && edge[i].flow > 0) {
dist[v] = dist[u] + 1;
if (v == T) return true;
que.push(v);
}
}
}
return false;
}
int cur[maxn];
int dfs(int u, int low) {
if (u == T) return low;
for (int &i = cur[u]; ~i; i = edge[i].next) {
int v = edge[i].to, flow = edge[i].flow;
if (dist[v] != dist[u] + 1 || flow <= 0) continue;
int tmp = dfs(v, std::min(flow, low));
if (tmp > 0) {
edge[i].flow -= tmp;
edge[i ^ 1].flow += tmp;
return tmp;
}
}
return 0;
}
int dinic() {
int ans = 0, tmp;
while (bfs(S, T)) {
while ((tmp = dfs(S, 0x3f3f3f3f)) > 0) ans += tmp;
}
return ans;
}
void init() {
cnt_edge = 0;
}
int main() {
//    freopen("in.txt", "r", stdin);
init();
scanf("%d", &n);
S = 0, T = n + 1;
for (int i = 1, val, cost, m, x; i <= n; i++) {
scanf("%d%d", &val, &cost);
if (val > cost) sum += val - cost, addedge(S, i, val - cost);
else addedge(i, T, cost - val);
scanf("%d", &m);
while (m--) {
scanf("%d", &x);
}