题目链接：

hdu:

bc:

wyh2000 and pupil

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 1304    Accepted Submission(s): 418

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

 

题解：

有解条件：每个连通分量都为二分图，且能够分为人数都大于1的两组。

先用黑白染色法来判断每个连通分量是否都为二分图，之后对每个联通分量，分成黑白两组后人数多的进第一组，人数少的进第二组。

这样贪心会有一个问题，第二个组的人数少于1，所以还要做一些调整。

代码：

#include
       
        #include
        
         #include
         
          #include
          
           using namespace std;const int maxn = 101010;const int INF = 0x3f3f3f3f;struct Edge {    int v, ne;    Edge(int v = 0, int ne = 0) :v(v), ne(ne) {};}egs[maxn * 2];int n, m;int head[maxn], tot;//vis[i]==-1:没访问过，==0：白色，==1：黑色int vis[maxn];void addEdge(int u, int v) {    egs[tot] = Edge(v, head[u]);    head[u] = tot++;}int vis2[maxn];//统计该连通分量黑白两色的人数void dfs2(int cur, int &cnt0, int &cnt1) {    vis2[cur] = 1;    if (vis[cur] == 0) cnt0++;    else cnt1++;    int p = head[cur];    while (p != -1) {        Edge &e = egs[p];        if (!vis2[e.v]) {            dfs2(e.v, cnt0, cnt1);        }        p = e.ne;    }}//二染色判二分图bool dfs(int cur) {    int p = head[cur];    while (p != -1) {        Edge &e = egs[p];        if (vis[e.v] == -1) {            vis[e.v] = vis[cur] ^ 1;            dfs(e.v);        }        else if (vis[e.v] == vis[cur]) {            return false;        }        p = e.ne;    }    return true;}void init() {    tot = 0;    memset(head, -1, sizeof(head));    memset(vis, -1, sizeof(vis));    memset(vis2, 0, sizeof(vis2));}int main() {    int tc;    scanf("%d", &tc);    while (tc--) {        init();        scanf("%d%d", &n, &m);        for (int i = 0; i < m; i++) {            int u, v;            scanf("%d%d", &u, &v);            u--, v--;            addEdge(u, v);            addEdge(v, u);        }        int ans = 0;        int flag = 0, adj = INF;        for (int i = 0; i < n; i++) {            if (vis[i] == -1) {                vis[i] = 0;                if (!dfs(i)) {                    flag = 1; break;                }                else {                    int cnt0 = 0, cnt1 = 0;                    dfs2(i, cnt0, cnt1);                    ans += max(cnt0, cnt1);                    if (abs(cnt0 - cnt1) > 0) {                        adj = min(adj, abs(cnt0 - cnt1));                    }                }            }        }        //调整        if (adj == INF) adj = 0;        if (n - ans < 1) { ans -= adj; }        if (flag || n - ans<1 || ans<1) {            printf("Poor wyh\n");        }        else {            printf("%d %d\n", max(ans, n - ans), min(ans, n - ans));        }    }    return 0;}