题意：n(n <= 20)个项目，m(m <= 50)个技术问题，做完一个项目可以有收益profit (<= 1000)，做完一个项目必须解决相应的技术问题，解决一个技术问题需要付出cost ( <= 1000)，技术问题之间有先后依赖关系，求最大收益。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4971
——>>项目必须解决相应的技术问题，技术问题之间也存在依赖，对应闭合图，最大收益对应最大权和。。于是，最大权闭合图，最小割，最大流上场。。
建图：
1）超级源S = n + m, 超级汇T = n + m + 1
2）对于每个项目i：S -> i (profit[i])
3）对于每个技术问题i：i + n -> T (cost[i])
4）对于项目 i 必须解决的技术问题 j：i -> j + n (INF)
5）对于技术问题 j 必须先解决的技术问题 i： i + n -> j + n (INF) （这里我觉得应为：j + n -> i + n (INF)，这样理解才对，可是对不上样例，提交也WA。。）
然后，Dinic上场。。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>using std::queue;
using std::min;const int MAXN = 20 + 50 + 10;
const int MAXM = 20 + 1000 + 2500 + 50 + 10;
const int INF = 0x3f3f3f3f;int n, m, sum;
int hed[MAXN];
int cur[MAXN], h[MAXN];
int ecnt;
int S, T;struct EDGE
{int to;int cap;int flow;int nxt;
} edges[MAXM << 1];void Init()
{ecnt = 0;memset(hed, -1, sizeof(hed));sum = 0;
}void AddEdge(int u, int v, int cap)
{edges[ecnt].to = v;edges[ecnt].cap = cap;edges[ecnt].flow = 0;edges[ecnt].nxt = hed[u];hed[u] = ecnt++;edges[ecnt].to = u;edges[ecnt].cap = 0;edges[ecnt].flow = 0;edges[ecnt].nxt = hed[v];hed[v] = ecnt++;
}void Read()
{int profit, cost, pc, tp;scanf("%d%d", &n, &m);S = n + m;T = n + m + 3;for (int i = 0; i < n; ++i){scanf("%d", &profit);AddEdge(S, i, profit);sum += profit;}for (int i = 0; i < m; ++i){scanf("%d", &cost);AddEdge(i + n, T, cost);}for (int i = 0; i < n; ++i){scanf("%d", &pc);for (int j = 0; j < pc; ++j){scanf("%d", &tp);AddEdge(i, tp + n, INF);}}for (int i = 0; i < m; ++i){for (int j = 0; j < m; ++j){scanf("%d", &tp);if (tp){AddEdge(i + n, j + n, INF);}}}
}bool Bfs()
{memset(h, -1, sizeof(h));queue<int> qu;qu.push(S);h[S] = 0;while (!qu.empty()){int u = qu.front();qu.pop();for (int e = hed[u]; e != -1; e = edges[e].nxt){int v = edges[e].to;if (h[v] == -1 && edges[e].cap > edges[e].flow){h[v] = h[u] + 1;qu.push(v);}}}return h[T] != -1;
}int Dfs(int u, int cap)
{if (u == T || cap == 0) return cap;int flow = 0, subFlow;for (int e = cur[u]; e != -1; e = edges[e].nxt){cur[u] = e;int v = edges[e].to;if (h[v] == h[u] + 1 && (subFlow = Dfs(v, min(cap, edges[e].cap - edges[e].flow))) > 0){flow += subFlow;edges[e].flow += subFlow;edges[e ^ 1].flow -= subFlow;cap -= subFlow;if (cap == 0) break;}}return flow;
}int Dinic()
{int maxFlow = 0;while (Bfs()){memcpy(cur, hed, sizeof(hed));maxFlow += Dfs(S, INF);}return maxFlow;
}int main()
{int t, kase = 0;scanf("%d", &t);while (t--){Init();Read();printf("Case #%d: %d\n", ++kase, sum - Dinic());}return 0;
}