POJ2239 Selecting Courses（二分图最大匹配）

题目链接

N节课，每节课在一个星期中的某一节，求最多能选几节课

好吧，想了半天没想出来，最后看了题解是二分图最大匹配，好弱

建图： 每节课 与 时间有一条边

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 using namespace std;
 8 const int N = 300 + 5;
 9 int vis[N], Left[N];
10 vector<int> g[N];
11 int n;
12 bool mach(int u)
13 {
14     int Size = g[u].size();
15     for (int i = 0; i < Size; i++)
16     {
17         int v = g[u][i];
18         if (!vis[v])
19         {
20             vis[v] = 1;
21             if (Left[v] == -1 || mach(Left[v]))
22             {
23                 Left[v] = u;
24                 return true;
25             }
26         }
27     }
28     return false;
29 }
30 int main()
31 {
32     while (scanf("%d", &n) != EOF)
33     {
34         int t, p, q;
35         for (int i = 0; i <= n; i++)
36             g[i].clear();
37         for (int i = 1; i <= n; i++)
38         {
39             scanf("%d", &t);
40             for (int j = 1; j <= t; j++)
41             {
42                 scanf("%d%d", &p, &q);
43                 g[i].push_back(p * 12 + q); // 可以将每一节转化成一个整数
44             }
45         }
46         memset(Left, -1, sizeof(Left));
47         int ans = 0;
48         for (int i = 1; i <= n; i++)
49         {
50             memset(vis, 0, sizeof(vis));
51             if (mach(i))
52                 ans++;
53         }
54         printf("%d\n", ans);
55     }
56     return 0;
57 }