ACM: HDU 5285 wyh2000 and pupil-二分图判定

 HDU 5285  wyh2000 and pupil

Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

 

Description

Young theoretical computer scientist wyh2000 is teaching his pupils. 

Wyh2000 has n pupils.Id of them are from  to .In order to increase the cohesion between pupils,wyh2000 decide to divide them into 2 groups.Each group has at least 1 pupil. 

Now that some pupils don't know each other(if  doesn't know ,then  doesn't know ).Wyh2000 hopes that if two pupils are in the same group,then they know each other,and the pupils of the first group must be as much as possible. 

Please help wyh2000 determine the pupils of first group and second group. If there is no solution, print "Poor wyh".

Input

In the first line, there is an integer  indicates the number of test cases. 

For each case, the first line contains two integers  indicate the number of pupil and the number of pupils don't konw each other. 

In the next m lines,each line contains 2 intergers <,indicates that  don't know  and  don't know ,the pair  will only appear once. 

Output

For each case, output the answer.

Sample Input

2
8 5
3 4
5 6
1 2
5 8
3 5
5 4
2 3
4 5
3 4
2 4

Sample Output

5 3
Poor wyh

/*/
二分图染色法判定

题解：

把两个相连的点标记颜色为不同颜色1和0，表示两位同学相互不认识；

统计两个颜色的个数。

注意，应为可能出现多个联通块的情况，这时候只要将各个联通块的最少的颜色

加起来，用总人数减去这个和就是人数最多那个组的人数了。

本题还要注意几个特判。 

AC代码： 
/*/

#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"string"
#include"cstdio"
#include"vector"
#include"cmath"
#include"queue"
using namespace std;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define MX 500005

struct Edge {
	int v,nxt;
} E[MX*2];

int col[3];

int Head[MX],erear;

void edge_init() {
	erear=0;
	memset(E,0);
	memset(Head,-1);
}
void edge_add(int u,int v) {
	E[erear].v=v;
	E[erear].nxt=Head[u];
	Head[u]=erear++;
}

int color[MX];
bool DFS(int u,int c) {
	color[u]=c;
	if(color[u]==1) {
		col[1]++;
	} else col[0]++;
	for(int i=Head[u]; ~i; i=E[i].nxt) {
		int v=E[i].v;
		if(color[v]!=-1&&color[u]==color[v]) return 0;
		else if(color[v]==-1) {
			if(DFS(v,c^1)==0) return 0;
		}
	}
	return 1;
}

int main() {
	int n,m;
	int T;
	cin>>T;
	while(T--) {
		scanf("%d%d",&n,&m);
		edge_init();
		for(int i=1; i<=m; i++) {
			int u,v;
			scanf("%d%d",&u,&v);
			edge_add(u,v);
			edge_add(v,u);
		}
		if(n<2) {
			puts("Poor wyh");
			continue;
		}
		if(!m) {
			printf("%d 1
",n-1);
		}
		memset(color,-1);
		bool sign=1;
		int minn=0;
		for(int i=1; i<=n; i++) {
			if(color[i]==-1) {
				memset(col,0);
				sign=DFS(i,0);
				minn+=min(col[0],col[1]);
				if(!sign)break;
			}

		}
		if(!sign)puts("Poor wyh") ;
		else printf("%d %d
",n-minn,minn);
	}
	return 0;
}