HDU 5285 wyh2000 and pupil(dfs或种类并查集)

wyh2000 and pupil

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 755    Accepted Submission(s): 251

Problem Description

Young theoretical computer scientist wyh2000 is teaching his pupils.

Wyh2000 has n pupils.Id of them are from 1 to n.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 a doesn't know b,then b doesn't know a).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 T indicates the number of test cases.

For each case, the first line contains two integers n,m 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 x,y(x<y),indicates that x don't know y and y don't know x,the pair (x,y) will only appear once.

T≤10,0≤n,m≤100000

 

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

 

大致题意：

n个点分成两组，m条边，每条边连接的两个点必须是在不同的两组，且第一组要尽量的大

思路：显然。dfs染色，不是二分图就无解。

还能够用种类并查集来做，把全部可能性合并，由对称性可知，仅仅统计1~n是祖先时。看祖先的子节点中黑白节点是多少个就ok

//468MS 4036K 1981 B C++  
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<ll,ll> pii;


const int N = 1e5+100;

int sz[N*2];
int sz2[N*2];
int fa[N*2];
int n,m;
void ini(){
        REP(i,2*n) fa[i] = i;
        REP(i,2*n) sz[i] = (i <= n);
        REP(i,2*n) sz2[i] = (i > n);
}
int getf(int x){
        return x == fa[x] ? x : fa[x] = getf(fa[x]);
}
bool same(int a,int b){
        return getf(a) == getf(b);
}
void Merge(int a,int b){
        int f1 = getf(a), f2 = getf(b);
        if(f1 == f2) return ;
        fa[f1] = f2;
        sz[f2] += sz[f1];
        sz2[f2] += sz2[f1];
}
int main(){
        int T;
        cin>>T;
        while(T--){
                scanf("%d%d",&n,&m);
                ini();
                bool flag = 0;
                REP(i,m){
                        int a,b;
                        scanf("%d%d",&a,&b);
                        if(flag) continue;
                        if(same(a,b) || same(a+n,b+n) ) flag = 1;
                        else Merge(a+n,b),Merge(a,b+n);
                }
                if( n < 2 || flag) puts("Poor wyh");
                else if(m == 0) printf("%d 1
",n-1);
                else {
                        int ans = 0;
                        REP(i,n){
                                if( fa[i] == i) ans += min(sz[i],sz2[i]);
                        }
                        printf("%d %d
",n-ans,ans);
                }

        }
}