【二分图】The Accomodation of Students

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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

Solution

交叉染色法判断是否为二分图，dfs实现匈牙利算法求二分图的最大匹配。

二分图相关概念：http://www.renfei.org/blog/bipartite-matching.html

趣写匈牙利算法：http://blog.csdn.net/dark_scope/article/details/8880547

Code

 1 #include<queue>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<iostream>
 5 using namespace std;
 6 
 7 const int A=40000*2+10,B=210;
 8 int s,n,m,x,y,t,k;
 9 int head[B],ex[B],check[B],mch[B];
10 struct edge{
11     int t,n;
12 }e[A];
13 bool flag=1;
14 queue<int>q;
15 
16 void build(int f,int tt)
17 {
18     e[++s].t=tt;
19     e[s].n=head[f];
20     head[f]=s;
21 }
22 
23 bool bfs()
24 {    
25     while(!q.empty())q.pop();
26     q.push(1);
27     ex[1]=1;
28     while(!q.empty())
29     {
30         t=q.front();
31         q.pop();
32         for(int i=head[t];i;i=e[i].n)
33         {
34             if(ex[e[i].t]==-1)
35             {
36                 ex[e[i].t]=!ex[t];
37                 q.push(e[i].t);
38             }
39             else
40             if(ex[e[i].t]==ex[t])
41                 return false;
42         }
43     }
44     return true;
45 }
46 
47 bool dfs(int x)
48 {
49     for(int i=head[x];i;i=e[i].n)
50     {
51         k=e[i].t;
52         if(!check[k])
53         {
54             check[k]=1;
55             if(mch[k]==-1||dfs(mch[k]))
56             {
57                 mch[k]=x;
58                 return true;
59             }
60         }
61     }
62     return false;
63 }
64 
65 void hungerian()
66 {
67     int ans=0;
68     for(int i=0;i<=n;++i)mch[i]=-1;
69     for(int i=1;i<=n;++i)
70     {
71         memset(check,0,sizeof(check));
72         if(dfs(i))++ans;
73     }
74     printf("%d\n",ans/2);//无向图
75 }
76 
77 int main()
78 {
79     while(scanf("%d%d",&n,&m)==2)
80     {
81         if(n==1)
82         {printf("No\n");continue;}//极端情况
83         s=0;flag=1;
84         memset(head,0,sizeof(head));//重要的初始化
85         for(int i=1;i<=n;++i)ex[i]=-1;
86         for(int i=1;i<=m;++i)
87         {
88             scanf("%d%d",&x,&y);
89             build(x,y);
90             build(y,x);
91         }
92         if(!bfs())printf("No\n");
93         else hungerian();
94     }
95     return 0;
96 }

Source

HDU 2444

posted @ 2017-02-06 20:07  Etta19  阅读(347)  评论(0)  编辑  收藏  举报