2018 ACM/ICPC 南京 I题 Magic Potion

题解：最大流板题；增加两个源点，一个汇点。第一个源点到第二个源点连边，权为K，然后第一个源点再连其他点（英雄点）边权各为1，然后英雄和怪物之间按照所给连边（边权为1）。

每个怪物连终点，边权为1；

参考代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define INF 0x3f3f3f3f
 4 const int maxn = 2100;
 5 int n,m,k,s,t,u,v,w,num,num1;
 6 struct Edge {
 7     int from, to, cap, flow;
 8 };
 9 vector<Edge> edges;
10 vector<int> G[maxn];
11 bool vis[maxn];
12 int d[maxn], cur[maxn];
13 void Init()
14 {
15     memset(d,0,sizeof d);
16     for(int i=0;i<=n+m+4;i++) G[i].clear();
17 }
18 void addedge(int from, int to, int cap)
19 {
20     edges.push_back((Edge){from, to, cap, 0});
21     edges.push_back((Edge){to, from, 0, 0});
22     int m = edges.size();
23     G[from].push_back(m-2); G[to].push_back(m-1);
24 }
25 bool bfs()
26 {
27     memset(vis,0,sizeof vis);
28     queue<int> q;
29     q.push(s);
30     d[s] = 0; vis[s] = 1;
31     while (!q.empty())
32     {
33         int x = q.front(); q.pop();
34         for(int i = 0; i < G[x].size(); ++i)
35         {
36             Edge &e = edges[G[x][i]];
37             if (!vis[e.to] && e.cap > e.flow)
38             {
39                 vis[e.to] = 1;
40                 d[e.to] = d[x] + 1;
41                 q.push(e.to);
42             }
43         }
44     }
45     return vis[t];
46 }
47
48 int dfs(int x,int a)
49 {
50     if(x == t || a == 0) return a;
51     int flow = 0, f;
52     for(int &i = cur[x]; i < G[x].size(); ++i)
53     {
54         Edge &e = edges[G[x][i]];
55         if (d[e.to] == d[x] + 1 && (f=dfs(e.to, min(a, e.cap-e.flow))) > 0)
56         {
57             e.flow += f;
58             edges[G[x][i]^1].flow -= f;
59             flow += f; a -= f;
60             if (a == 0) break;
61         }
62     }
63     return flow;
64 }
65
66 int Maxflow(int s, int t)
67 {
68     int flow = 0;
69     while (bfs())
70     {
71         memset(cur,0,sizeof cur);
72         flow += dfs(s, INF);
73     }
74     return flow;
75 }
76 int main()
77 {
78     scanf("%d%d%d",&n,&m,&k);
79     Init();s=1;t=n+m+3;addedge(s,2,k);
80     for(int i=1;i<=n;++i) addedge(s,i+2,1),addedge(2,i+2,1);
81     for(int i=1;i<=n;++i)
82     {
83         scanf("%d",&num);
84         while(num--)
85         {
86             scanf("%d",&num1);
87             addedge(i+2,num1+n+2,1);
88         }
89     }
90     for(int i=1;i<=m;++i) addedge(i+n+2,t,1);
91     printf("%d\n",Maxflow(s,t));
92     return 0;
93  } 

原文地址：https://www.cnblogs.com/songorz/p/10331465.html