二分图判定+点染色/并查集 BestCoder Round #48 ($) 1002 wyh2000 and pupil

题目传送门

/*
    二分图判定+点染色：因为有很多联通块，要对所有点二分图匹配，若不能，存在点是无法分配的，no
            每一次二分图匹配时，将点多的集合加大最后第一个集合去
    注意：n <= 1，no，两个集合都至少有一人；ans == n，扔一个给另一个集合
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <map>
using namespace std;

const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
vector<int> G[MAXN];
int col[MAXN];
int cnt[3];
int n, m;

bool DFS(int u)
{
    for (int i=0; i<G[u].size (); ++i)
    {
        int v = G[u][i];
        if (col[v] == -1) {
            col[v] = 3 - col[u];    cnt[col[v]]++;
            if (!DFS (v))   return false;
        }
        else if (col[v] == col[u])  return false;
    }

    return true;
}

void work(void)
{
    memset (col, -1, sizeof (col));
    int ans = 0;
    for (int i=1; i<=n; ++i)
    {
        if (col[i] == -1)
        {
            col[i] = 1; cnt[1] = 1; cnt[2] = 0;
            if (!DFS (i)) {
                puts ("Poor wyh");  return ;
            }
            ans += max (cnt[1], cnt[2]);
        }
    }
    if (ans == n)   ans--;
    printf ("%d %d
", ans, n - ans);
}

int main(void)      //BestCoder Round #48 ($) 1002    wyh2000 and pupil
{
    int t;   scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d%d", &n, &m);
        if (n <= 1) {
            puts ("Poor wyh");  continue;
        }
        for (int i=1; i<=n; ++i)    G[i].clear ();
        for (int i=1; i<=m; ++i)
        {
            int u, v;   scanf ("%d%d", &u, &v);
            G[u].push_back (v); G[v].push_back (u);
        }
        work ();
    }
    

    return 0;
}

/*
    并查集：有点像POJ食物链的做法， 分为(1, n)和(n+1, 2*n)两个集合，两种情况都试一下。
            rt[i] == -1表示是该集合的根，用sz数组记录集合的大小
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <map>
using namespace std;

const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n, m;
struct UF   {
    int rt[MAXN*2], sz[MAXN*2], sz2[MAXN*2];
    void init(void) {
        memset (rt, -1, sizeof (rt));
        for (int i=1; i<=2*n; ++i)  sz[i] = (i <= n), sz2[i] = (i > n);
    }
    int Find(int x) {
        return rt[x] == -1 ? x : rt[x] = Find (rt[x]); 
    }
    void Union(int x, int y)    {
        x = Find (x);   y = Find (y);
        if (x == y) return ;
        rt[x] = y;
        sz[y] += sz[x]; sz2[y] += sz2[x];
    }
    bool same(int x, int y) {
        return Find (x) == Find (y);
    }
}uf;

int main(void)      //BestCoder Round #48 ($) 1002    wyh2000 and pupil
{
    int t;   scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d%d", &n, &m);
        bool ok = true; uf.init ();
        for (int i=1; i<=m; ++i)
        {
            int u, v;   scanf ("%d%d", &u, &v);
            if (!ok)   continue;
            if (uf.same (u, v) || uf.same (u+n, v+n))   ok = false;
            else    uf.Union (u, v+n), uf.Union (u+n, v);
        }

        if (!ok || n <= 1)    puts ("Poor wyh");
        else if (m == 0)    printf ("%d 1
", n - 1);
        else    {
            int ans = 0;
            for (int i=1; i<=n; ++i)    {
                if (uf.rt[i] == -1) {
                    ans += min (uf.sz[i], uf.sz2[i]);
                }
            }
            printf ("%d %d
", n - ans, ans);
        }
    }

    return 0;
}