HDU2444_The Accomodation of Students(二分图/最大匹配+判二分图)

解题报告

题目传送门

题意：

问学生是否能分成两部分，每一部分的人都不相认识，如果能分成的话，两两认识的人可以去开房哈。求最大的开房数。

思路：

先判断是否能生成二分图，用黑白标记法，如果颜色相同就没办法分成两部分。

再求二分图的最大匹配。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
int mmap[310][310],vis[310],pre[310],n,m,_hash[300];
int dfs(int x) {
for(int i=1; i<=n; i++) {
if(!vis[i]&&mmap[x][i]) {
vis[i]=1;
if(pre[i]==-1||dfs(pre[i])) {
pre[i]=x;
return 1;
}
}
}
return 0;
}
int main() {
int t,i,j,u,v;
while(~scanf("%d%d",&n,&m)) {
int f=0,cnt=0;
memset(pre,-1,sizeof(pre));
memset(mmap,0,sizeof(mmap));
memset(_hash,0,sizeof(_hash));
for(i=0; i<m; i++) {
scanf("%d%d",&u,&v);
if((_hash[u]==_hash[v])&&_hash[u]!=0)
f=1;
if(!_hash[u]&&!_hash[v]) {
_hash[u]=1;
_hash[v]=2;
}
if(!_hash[u]&&_hash[v]) {
if(_hash[v]==1)
_hash[u]=2;
else _hash[u]=1;
}
if(!_hash[v]&&_hash[u]) {
if(_hash[u]==1)
_hash[v]=2;
else _hash[v]=1;
}
mmap[u][v]=mmap[v][u]=1;
}
if(f) {
printf("No\n");
} else {
int ans=0;
for(i=1; i<=n; i++) {
memset(vis,0,sizeof(vis));
ans+=dfs(i);
}
printf("%d\n",ans/2);
}
}
return 0;
}

The Accomodation of Students

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2529    Accepted Submission(s): 1198


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

 

Source

2008 Asia Harbin Regional Contest Online