HDU5285：wyh2000 and pupil（二分图染色 & 贪心）

wyh2000 and pupil

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1824    Accepted Submission(s): 599

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
  

 

Source

BestCoder Round #48 ($)

题意：N个人，有M对人互不认识，现在要将他们分成A，B两堆人，要求两堆人各至少有一个人，且同一堆里面的人要互相认识，A堆的人数要尽量多，输出两堆的人数。

思路：基础的二分图染色，首先是二分图的判定，出现奇数环就表示无法分配，否则取中max(黑点数，白点数)，要每堆至少有一人，如果答案A堆有N个人，说明必定所有点都是独立点，即M=0。此时分1个人给B堆就行了。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <vector>
# define pb push_back
using namespace std;
const int maxn = 1e5+3;
vector<int>g[maxn];
int col[maxn], cnt, tot, ans, flag;
void dfs(int cur, int co)
{
    if(flag) return;
    if(co == 1) ++cnt; ++tot;
    col[cur] = co;
    for(int i=0; i<g[cur].size(); ++i)
    {
        if(col[g[cur][i]] == co)
        {
            flag = 1;
            return;
        }
        else if(col[g[cur][i]] == -1)
            dfs(g[cur][i], co^1);
    }
}
int main()
{
    int T, x, y, n, m;
    scanf("%d",&T);
    while(T--)
    {
        cnt = ans = flag = 0;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; ++i) g[i].clear(), col[i] = -1;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            g[x].pb(y);
            g[y].pb(x);
        }
        if(n == 0 || n == 1)
        {
            puts("Poor wyh");
            continue;
        }
        for(int i=1; i<=n; ++i)
        {
            if(col[i] != -1) continue;
            tot=cnt=0;
            dfs(i, 0);
            if(flag) break;
            ans += max(cnt, tot-cnt);
        }
        if(flag) puts("Poor wyh");
        else
        {
            if(ans != n) printf("%d %d\n",ans,n-ans);
            else printf("%d %d\n",ans-1, 1);
        }
    }
    return 0;
}