The Accomodation of Students
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7367 Accepted Submission(s): 3287
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：n个点，m条边，判断是否是二部图，如果不是输出NO,如果是，输出最大匹配数
首先用染色法判断二部图，先找一个图中的点加入队列，将与其相连的点都变成与他相反的颜色，然后把这些点加入队列中，如果出现相连的两点颜色相同，那么就不是二部图，根据二部图的定义，很容易就可以明白
如果是二部图，用匈牙利算法找出最大匹配，由于建图是双向，如果u和v可以匹配，那么v和u也可以产生一个匹配，所以最后得到的匹配数为所求答案的二倍，除2即可
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=205;
int n,m,match[N],pre;
bool vis[N];
vector<int>g[N];
int color[N];
int find(int u)
{
for(int i=0; i<g[u].size(); i++)
{
if(!vis[g[u][i]])
{
vis[g[u][i]]=1;
if(!match[g[u][i]]||find(match[g[u][i]]))
{
match[g[u][i]]=u;
return 1;
}
}
}
return 0;
}
int matching()
{
int ans=0;
mem(match,0);
for(int i=1;i<=n;i++)
{
mem(vis,0);
ans+=find(i);
}
return ans;
}
bool check()
{
queue<int>q;
q.push(pre);
color[pre]=1;
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=0;i<g[u].size();i++)
{
int v=g[u][i];
if(color[v]==-1)
{
color[v]=-color[u];
q.push(v);
}
else if(color[v]==color[u])
return 0;
}
}
return 1;
}
int main()
{
int x,y;
while(~scanf("%d%d",&n,&m))
{
mem(g,0);
mem(color,-1);
pre=0;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
if(!pre)pre=x;
g[x].push_back(y);
g[y].push_back(x);
}
if(!check())
puts("No");
else
printf("%d\n",matching()/2);
}
return 0;
}
