[CEOI2008]order

嘟嘟嘟

最小割。

不过割掉边代表付出某一种代价，而题中让求的是利润。所以我们先把完成所有任务的总利润加起来，然后算没有完成任务付出的代价。

建图比较明白：

1.源点向任务连一条边，容量为完成这个任务赚到的钱。割掉代表这个任务没完成。

2.每一个机器向汇点连一条边，容量为买这个机器所花的钱。割掉代表没买这个机器。

3.每一个任务向完成这个任务要用到的机器连边，容量为租用该机器所花的钱。

图大概是这样的：

1 #include<cstdio>
2 #include<iostream>
3 #include<cmath>
4 #include<algorithm>
5 #include<cstring>
6 #include<cstdlib>
7 #include<cctype>
8 #include<vector>
9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(' ')
14 #define Mem(a, x) memset(a, x, sizeof(a))
15 #define rg register
16 typedef long long ll;
17 typedef double db;
18 const int INF = 0x3f3f3f3f;
19 const db eps = 1e-8;
20 const int maxn = 2.5e3 + 5;
21 inline ll read()
22 {
23     ll ans = 0;
24     char ch = getchar(), last = ' ';
25     while(!isdigit(ch)) {last = ch; ch = getchar();}
26     while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
27     if(last == '-') ans = -ans;
28     return ans;
29 }
30 inline void write(ll x)
31 {
32     if(x < 0) x = -x, putchar('-');
33     if(x >= 10) write(x / 10);
34     putchar(x % 10 + '0');
35 }
36
37 int n, m, t;
38 int sum = 0;
39
40 struct Edge
41 {
42     int from, to, cap, flow;
43 };
44 vector<Edge> edges;
45 vector<int> G[maxn];
46 void addEdge(int from, int to, int w)
47 {
48     edges.push_back((Edge){from, to, w, 0});
49     edges.push_back((Edge){to, from, 0, 0});
50     int sz = edges.size();
51     G[from].push_back(sz - 2);
52     G[to].push_back(sz - 1);
53 }
54
55 int dis[maxn];
56 bool bfs()
57 {
58     Mem(dis, 0); dis[0] = 1;
59     queue<int> q; q.push(0);
60     while(!q.empty())
61     {
62         int now = q.front(); q.pop();
63         for(int i = 0; i < (int)G[now].size(); ++i)
64         {
65             Edge& e = edges[G[now][i]];
66             if(!dis[e.to] && e.cap > e.flow)
67             {
68                 dis[e.to] = dis[now] + 1;
69                 q.push(e.to);
70             }
71         }
72     }
73     return dis[t];
74 }
75 int cur[maxn];
76 int dfs(int now, int res)
77 {
78     if(now == t || res == 0) return res;
79     int flow = 0, f;
80     for(int& i = cur[now]; i < (int)G[now].size(); ++i)
81     {
82         Edge& e = edges[G[now][i]];
83         if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
84         {
85             e.flow += f;
86             edges[G[now][i] ^ 1].flow -= f;
87             flow += f; res -= f;
88             if(res == 0) break;
89         }
90     }
91     return flow;
92 }
93
94 int minCut()
95 {
96     int flow = 0;
97     while(bfs())
98     {
99         Mem(cur, 0);
100         flow += dfs(0, INF);
101     }
102     return flow;
103 }
104
105 int main()
106 {
107     n = read(); m = read();
108     t = n + m + 1;
109     for(int i = 1; i <= n; ++i)
110     {
111         int x = read(), d = read();
112         sum += x; addEdge(0, i, x);
113         for(int j = 1; j <= d; ++j)
114         {
115             int y = read(), w = read();
116             addEdge(i, n + y, w);
117         }
118     }
119     for(int i = 1; i <= m; ++i)
120     {
121         int w = read();
122         addEdge(n + i, t, w);
123     }
124     write(sum - minCut()); enter;
125     return 0;
126 }