2018 AICCSA Programming Contest

思路:如果存在大于0的交面积的话, 那么肯定能找到一条水平的直线和一条垂直的直线,

使得水平直线的左右两边点的个数相等且为n, 垂直直线的左右两边点的个数相等且为n

也就是说不能有点在这两条线上, 否则交面积为0

然后左上角的点和右下角的点配对, 左下角的点和右上角的点配对

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 2e5 + 5, M = 1e5 + 5;
const int MOD = 1e9 + 7;
pii a[N];
int fac[M];
bool cmp(pii a, pii b) {
    return a.se < b.se;
}
void init() {
    fac[0] = 1;
    for (int i = 1; i < M; i++) fac[i] = (1LL * fac[i-1] * i) % MOD;
}
int main() {
    int T, n;
    init();
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for (int i = 1; i <= 2*n; i++) scanf("%d %d", &a[i].fi, &a[i].se);
        bool f = false;
        sort(a+1, a+1+2*n);
        double x = 0, y = 0;
        if(a[n].fi != a[n+1].fi) x = (a[n].fi + a[n+1].fi) / 2.0;
        else f = true;

        sort(a+1, a+1+2*n, cmp);
        if(a[n].se != a[n+1].se) y = (a[n].se + a[n+1].se) / 2.0;
        else f = true;
        int cnt = 0;
        for (int i = 1; i <= 2*n; i++) if(a[i].fi > x && a[i].se > y) cnt++;
        if(f) printf("0\n");
        else printf("%lld\n", (1LL * fac[cnt] * fac[n-cnt]) % MOD);
    }
    return 0;
}

View Code