The Accomodation of Students
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8418 Accepted Submission(s): 3709
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.
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
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6
Sample Output
No
3
3
Source
Recommend
gaojie
先判断能不能分成二分图 ， 不是就输出No。。只有1个的时候也不是
然后求完美匹配即可。。。我不会写匈牙利了。。。。只会写hk。。。还是套模板。。。
因为没有分左右 所以要除2
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = , INF = 0x7fffffff;
int dx[maxn], dy[maxn], cx[maxn], cy[maxn], used[maxn], vis[maxn];
int nx, ny, dis;
vector<int> G[];
int n, m;
int bfs()
{
queue<int> Q;
dis = INF;
mem(dx, -);
mem(dy, -);
for(int i=; i<=nx; i++)
{
if(cx[i] == -)
{
Q.push(i);
dx[i] = ;
}
}
while(!Q.empty())
{
int u = Q.front(); Q.pop();
if(dx[u] > dis) break;
for(int v=; v<G[u].size(); v++)
{
int i = G[u][v];
if(dy[i] == -)
{
dy[i] = dx[u] + ;
if(cy[i] == -) dis = cy[i];
else
{
dx[cy[i]] = dy[i] + ;
Q.push(cy[i]);
}
}
}
}
return dis != INF;
} int dfs(int u)
{
for(int v=; v<G[u].size(); v++)
{
int i=G[u][v];
if(!used[i] && dy[i] == dx[u] + )
{
used[i] = ;
if(cy[i] != - && dis == dy[i]) continue;
if(cy[i] == - || dfs(cy[i]))
{
cy[i] = u;
cx[u] = i;
return ;
}
}
}
return ;
} int hk()
{
int res = ;
mem(cx, -);
mem(cy, -);
while(bfs())
{
mem(used, );
for(int i=; i<=nx; i++)
if(cx[i] == - && dfs(i))
res++;
}
return res;
} int istwo(int u)
{
queue<int> E;
mem(vis, -);
E.push(u);
vis[u] = ;
while(!E.empty())
{
u = E.front(); E.pop();
for(int i=; i<G[u].size(); i++)
{
int v = G[u][i];
if(vis[v] == -)
{
if(vis[u] == ) vis[v] = ;
else vis[v] = ;
E.push(v);
}
else if(vis[v] == vis[u])
return ;
}
}
return ;
} int main()
{
while(cin>> n >> m && n+m)
{
for(int i=; i<maxn; i++) G[i].clear(); for(int i=; i<m; i++)
{
int u, v;
cin>> u >> v;
G[u].push_back(v);
G[v].push_back(u);
} if(!istwo() || n == )
{
cout<< "No" <<endl;
continue;
}
nx = n; ny = n;
cout<< hk()/ <<endl; } return ;
}