Kattis - flippingcards 【并查集】

题意

给出 N 对 数字 然后 每次从一对中 取出一个数字 判断 能否有一种取出的方案 取出的每个数字 都是不同的

思路

将每一对数字 连上一条边 然后 最后 判断每一个连通块里面 边的个数 是否 大于等于 点的个数 用并查集判断

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const double PI  = 3.14159265358979323846264338327;
const double E   = exp(1);
const double eps = 1e-6;

const int INF  = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int MOD  = 1e9 + 7;

int pre[maxn];

int find(int x)
{
    int r = x;
    while (pre[r] != r)
        r = pre[r];
    int j = x, i;
    while (j != r)
    {
        i = pre[j];
        pre[j] = r;
        j = i;
    }
    return r;
}

void join(int x, int y)
{
    int fx = find(x), fy = find(y);
    if (x != fy)
        pre[fx] = fy;
}   

int p[maxn], q[maxn];

struct Node
{
    map <int, int> M;
    int tot;
};

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        scanf("%d", &n);
        for (int i = 0; i < maxn; i++)
            pre[i] = i;
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &p[i], &q[i]);
            join(p[i], q[i]);
        }
        map <int, Node> m;
        int flag = 1;
        for (int i = 0; i < n; i++)
        {
            int num = find(p[i]);
            m[num].M[q[i]] = 1;
            m[num].M[p[i]] = 1;
            m[num].tot ++;
            if (m[num].M.size() < m[num].tot)
            {
                flag = 0;
                break;
            }
        }
        if (flag)
            printf("possible\n");
        else
            printf("impossible\n");
    }
}