Codeforces Round #440 (Div. 1, based on Technocup 2018 Elimination Round 2) C - Points, Lines and Ready-made Titles

C - Points, Lines and Ready-made Titles

把行列看成是图上的点， 一个点（x, y）就相当于x行 向 y列建立一条边， 我们能得出如果一个联通块是一棵树方案数是2 ^ n - 1

否则是2 ^ n。 各个联通块乘起来就是答案。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 4e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;

int n, x[N], y[N], hs[N], tot;
int bin[N], fa[N], ecnt[N], pcnt[N];

int getRoot(int x) {
    return x == fa[x] ? x : fa[x] = getRoot(fa[x]);
}

int main() {
    for(int i = bin[] = ; i < N; i++) bin[i] = bin[i - ] *  % mod;
    scanf("%d", &n);
    for(int i = ; i <= n; i++) {
        scanf("%d%d", &x[i], &y[i]);
        hs[++tot] = x[i];
        hs[++tot] = y[i];
    }
    sort(hs + , hs +  + tot);
    tot = unique(hs + , hs +  + tot) - hs - ;
    for(int i = ; i <= n; i++) {
        x[i] = lower_bound(hs + , hs +  + tot, x[i]) - hs;
        y[i] = lower_bound(hs + , hs +  + tot, y[i]) - hs;
    }
    for(int i = ; i <=  * tot; i++) fa[i] = i, ecnt[i] = , pcnt[i] = ;
    for(int i = ; i <= n; i++) {
        int X = getRoot(x[i]);
        int Y = getRoot(y[i] + tot);
        if(X == Y) {
            ecnt[X]++;
        } else {
            ecnt[X] += ecnt[Y] + ;
            pcnt[X] += pcnt[Y];
            fa[Y] = X;
        }
    }
    LL ans = ;
    for(int i = ; i <=  * tot; i++) {
        if(i != fa[i]) continue;
        if(ecnt[i] < pcnt[i]) ans = (ans * (bin[pcnt[i]] -  + mod) % mod) % mod;
        else ans = (ans * bin[pcnt[i]]) % mod;
    }
    printf("%lld\n", ans);
    return ;
}

/*
*/