The Accomodation of Students Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5451 Accepted Submis ...
The Accomodation of Students
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5451 Accepted Submission(s): 2491
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
題目大意：有n個學生，有m對人是認識的，每一對認識的人能分到一間房，問能否把n個學生分成兩部分，每部分內的學生互不認識，而兩部分之間的學生認識。如果可以分成兩部分，就算出房間最多需要多少間，否則就輸出No。
思路：二分判定用染色法，找最大匹配數用匈牙利演算法
代碼：
#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
const int MAX=1e5+5;
using namespace std;
vector<int>mp[MAX];
int d[MAX],n,m,vis[MAX],link[MAX];
int dfs(int x,int f) //染色法
{
d[x]=f;
for(int i=0; i<mp[x].size(); i++)
{
if(d[x]==d[mp[x][i]])
return 0;
int u=mp[x][i];
if(d[u]==0)
if(!dfs(u,-f))
return 0;
}
return 1;
}
int dfs2(int x) //找最大匹配
{
for(int i=0; i<mp[x].size(); i++)
{
int v=mp[x][i];
if(!vis[v])
{
vis[v]=1;
if(link[v]==-1||dfs2(link[v]))
{
link[v]=x;
return 1;
}
}
}
return 0;
}
int find() //找最大匹配
{
int res=0;
memset(link,-1,sizeof(link));
for(int i=1; i<=n; i++)
{
memset(vis,0,sizeof(vis));
if(dfs2(i))
res++;
}
return res;
}
int main()
{
int a,b;
while(cin>>n>>m)
{
for(int i=1; i<=n; i++)
if(mp[i].size())
mp[i].clear();
for(int i=0; i<m; i++)
{
scanf("%d%d",&a,&b);
mp[a].push_back(b);
mp[b].push_back(a);
}
int flag=1;
memset(d,0,sizeof(d));
for(int i=1; i<=n; i++) //染色
{
if(mp[i].size())
if(!d[i])
if(!dfs(i,1))
{
flag=0;
break;
}
}
if(!flag)
cout<<"No"<<endl;
else
{
cout<<find()/2<<endl;
}
}
}
