The Accomodation of Students(判断二分图以及求二分图最大匹配)
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
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
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
Sample Input
Sample Output
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
using namespace std;
int n,m;
int f[];
int A[],B[],match[],book[];
vector <int> V[];//邻接表储存边的关系
void init()
{
memset(match,,sizeof(match));
for (int i=;i<=n;i++)
V[i].clear();
for (int i=; i<=*n; i++)
f[i]=i;
}
int find(int x)
{
int r=x,i=x,t;
while (r!=f[r]) r=f[r];
while (i!=r)
{
t=f[i];
f[i]=r;
i=t;
}
return r;
}
void mix(int x,int y)
{
int fx=find(x),fy=find(y);
f[fx]=fy;
}
bool IsTwo()//染色法求二分图
{
memset(book,,sizeof(book));
queue <int> Q;
Q.push();
book[]=;
while (!Q.empty())
{
int temp=Q.front();
Q.pop();
for (int i=;i<V[temp].size();i++)
{
int num=V[temp][i];
if (book[num]==)
{
book[num]=-book[temp];
Q.push(num);
}
else if (book[num]==book[temp]) return ;
}
}
return ;
}
int dfs(int u)
{
for (int i=; i<V[u].size(); i++)
{
int pos=V[u][i];
if (book[pos]==)
{
book[pos]=;
if (match[pos]==||dfs(match[pos]))
{
match[pos]=u;
return ;
}
}
}
return ;
}
int solve()
{
int ans=;
for (int i=; i<=n; i++)
{
memset(book,,sizeof(book));
if (dfs(i)) ans++;
}
return ans;
}
int main()
{
int a,b;
while (scanf("%d%d",&n,&m)>)
{
init();
int ok=;
while (m--)
{
scanf("%d%d",&a,&b);
if (find(a)==find(b))
ok=;
if (ok)
{
mix(a,b+n);
mix(b,a+n);
V[a].push_back(b);
V[b].push_back(a);
}
}
if (!ok)
{
puts("No");
continue;
}
/*if (!IsTwo())
{
puts("No");
continue;
}*/
printf("%d\n",solve()/);
}
return ;
}
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