BNUOJ 12756 Social Holidaying（二分匹配）

題目連接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=12756

Social Holidaying

Time Limit: 3000ms

Memory Limit: 131072KB

 

This problem will be judged on UVALive. Original ID:  5874
64-bit integer IO format:  %lld      Java class name:  Main

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Source

Regionals 2011, Asia - Kuala Lumpur

 

題目大意：先輸入一個P表示有幾組測試數據，再輸入一個n，m分別表示A、B集合中分別有多少元素。在A集合中找到多少對數相加=B集合中的元素。A集合中的數不能夠重複，可是B集合中的數字能夠重複出現。

解題思路：二分匹配。先將A集合中的和的全部可能狀況列出來，而後在B集合中搜索便可。

 

詳見代碼。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int ss[9999999],Map[410][410];
 8 int ok[410],vis[410];
 9 int a[410],b[410];
10 int n,m;
11 
12 bool Find(int x)
13 {
14     for (int i=1; i<=n; i++)
15     {
16         if (!vis[i]&&Map[x][i]==1)
17         {
18             vis[i]=1;
19             if (!ok[i])
20             {
21                 ok[i]=x;
22                 return true;
23             }
24             else
25             {
26                 if (Find(ok[i]))
27                 {
28                     ok[i]=x;
29                     return true;
30                 }
31             }
32         }
33     }
34     return false;
35 }
36 
37 int main()
38 {
39     int t;
40     scanf("%d",&t);
41     while (t--)
42     {
43         scanf("%d%d",&n,&m);
44         int ans=0;
45         memset(vis,0,sizeof(vis));
46         memset(Map,0,sizeof(Map));
47         memset(ok,0,sizeof(ok));
48         memset(ss,0,sizeof(ss));
49         for (int i=1; i<=n; i++)
50         {
51             scanf("%d",&a[i]);
52         }
53         for (int i=1; i<=m; i++)
54         {
55             scanf("%d",&b[i]);
56             ss[b[i]]=1;
57         }
58         for (int i=1; i<=n; i++)
59         {
60             for (int j=i+1; j<=n; j++)
61             {
62                 if (ss[a[i]+a[j]]==1)
63                     Map[i][j]=Map[j][i]=1;
64             }
65         }
66         for (int i=1; i<=n; i++)
67         {
68             memset(vis,0,sizeof(vis));
69             if (Find(i))
70                 ans++;
71         }
72         printf ("%d\n",ans/2);
73     }
74     return 0;
75 }