首页 > 代码库 > hdu 2444 The Accomodation of Students

hdu 2444 The Accomodation of Students

2024-07-16 17:15:54   221人阅读

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2458    Accepted Submission(s): 1177


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
先判断 A->B ,B->C ,C->A 这种情况在不在，如果不存在再用匈牙利求出最大匹配
 1 #include<string> 2 #include<cstdio> 3 #include<iostream> 4 #include<vector> 5 #include<queue> 6 #include<stack> 7 #include<algorithm> 8 #include<cstring> 9 #include<stdlib.h>10 #include<string>11 #include<cmath>12 using namespace std;13 #define pb push_back14 #define ll __int6415 int fa[210],vit[210],p[210][210],x[210],y[210],cnt1,cnt2,cx[210];16 vector<int >ko[210];17 int n,m;18 int Find(int pos){19     for(int i=0;i<ko[pos].size();i++){20         int to=ko[pos][i];21         if(!vit[to]){22             vit[to]=1;23             if(!cx[to]||Find(cx[to])){24                 cx[to]=pos;25                 return 1;26             }27         }28     }29     return 0;30 }31 void solve(){32     int cnt=0;33     memset(cx,0,sizeof(cx));34     for(int i=1;i<=cnt1;i++){35         memset(vit,0,sizeof(vit));36         if(Find(x[i])) cnt++;37     }38     cout<<cnt<<endl;39 }40 int main(){41     while(cin>>n>>m){42         for(int i=1;i<=n;i++) ko[i].clear();43         memset(p,0,sizeof(p));44         memset(vit,0,sizeof(vit));45         for(int i=1;i<=m;i++){46             int a,b;scanf("%d%d",&a,&b);47             p[a][b]=1,p[b][a]=1;48             ko[a].pb(b);49             ko[b].pb(a);50         }51         for(int i=1;i<=n;i++)52         if(!vit[i]){53             vit[i]=1;54             fa[i]=0;55             queue<int >kk;56             kk.push(i);57             while(!kk.empty()){58                 int pos=kk.front();59                 kk.pop();60                 for(int j=0;j<ko[pos].size();j++){61                     int to=ko[pos][j];62                     if(!vit[to]){63                         fa[to]=!fa[pos];   //把相互认识的人放在不同的集合里里面64                         vit[to]=1;65                         kk.push(to);66                     }67                 }68             }69         }70         cnt1=cnt2=0;71         for(int i=1;i<=n;i++)72             if(!fa[i]) x[++cnt1]=i;73         for(int i=1;i<=n;i++)74             if(fa[i]) y[++cnt2]=i;75         int can=1;76         for(int i=1;i<=cnt1;i++)77             for(int j=1;j<=cnt1;j++)78                 if(p[x[i]][x[j]]) can=0;79           for(int i=1;i<=cnt2;i++)80             for(int j=1;j<=cnt2;j++)81                 if(p[y[i]][y[j]]) can=0;82         if(!can){83             cout<<"No"<<endl;84             continue;85         }86         else  solve();//不存在那种情况，进行匹配87     }88 }

 


联系
我们