HD2444The Accomodation of Students(并查集判断二分图+匹配)

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4091    Accepted Submission(s): 1876


Problem Description
There are a group of students. Some of them may know each other, while others don't. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don't know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

Calculate the maximum number of pairs that can be arranged into these double rooms.
 

 

Input
For each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

Proceed to the end of file.

 

 

Output
If these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
 

 

Sample Input
4 4 1 2 1 3 1 4 2 3 6 5 1 2 1 3 1 4 2 5 3 6
 

 

Sample Output
No 3
 
题意：A认识B,B认识C但是A不认识C,把ABC分成两个集合，每个集合中的人都认识，不能分成输出No,能的话然后从两个集合中选两个认识的组一个房间也就是在两个集合中求最大匹配，用到了种类并查集
 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cstdio>
 5 
 6 using namespace std;
 7 const int MAX = 210;
 8 int father[MAX],a[MAX];
 9 int g[MAX][MAX],vis[MAX],link[MAX];
10 int n,m;
11 int find_father(int x)
12 {
13     if(x == father[x])
14         return x;
15     int t = find_father(father[x]);
16     a[x] = (a[father[x]] + a[x]) % 2;
17     return father[x] = t;
18 }
19 int dfs(int x)
20 {
21     for(int i = 1; i <= n; i++)
22     {
23         if(vis[i] == 0 && g[x][i])
24         {
25             vis[i] = 1;
26             if(link[i] == 0 || dfs(link[i]))
27             {
28                 link[i] = x;
29                 return true;
30             }
31         }
32     }
33     return false;
34 }
35 int main()
36 {
37     while(scanf("%d%d",&n,&m) != EOF)
38     {
39         for(int i = 0; i <= n; i++)
40         {
41             father[i] = i;
42             a[i] = 0;
43         }
44         memset(g,0,sizeof(g));
45         memset(link,0,sizeof(link));
46         int flag = 0,ans = 0;
47         while(m--)
48         {
49             int x,y,fx,fy;
50             scanf("%d%d",&x,&y);
51             g[x][y] = g[y][x] = 1;
52             fx = find_father(x);
53             fy = find_father(y);
54             if(fx != fy)
55             {
56                 father[fy] = fx;
57                 if(a[x] == 0)
58                 {
59                     a[fy] = 1 - a[y];
60                 }
61                 else
62                     a[fy] = a[y];
63             }
64             else
65             {
66                 if(a[x] == a[y])
67                 {
68                     flag = 1;
69                 }
70             }
71         }
72         if(flag)
73         {
74             printf("No\n");
75         }
76         else
77         {
78             for(int i = 1; i <= n; i++)
79             {
80                 memset(vis,0,sizeof(vis));
81                 if(dfs(i))
82                     ans++;
83             }
84             printf("%d\n",ans/2);
85         }
86     }
87 
88     return 0;
89 }
View Code

 

posted @ 2015-12-01 23:33  zhaop  阅读(180)  评论(0)  编辑  收藏  举报