Educational Codeforces Round 68 E. Count The Rectangles

Educational Codeforces Round 68 E. Count The Rectangles

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题意:

给出不超过\(n,n\leq 5000\)条直线,问共形成多少个矩形。

思路:

考虑到\(n\)的范围不大,所以可以暴力枚举两条平行的直线,接下来处理的就是与其垂直的直线的数量。
满足相交成矩形有两个条件,假如我们枚举的直线是垂直于\(x\)轴的,那么两个条件即为\(low\leq y_i\leq high,x_{i,0}\leq left,right\leq x_{i,1}\)
所以我们可以考虑扫描线的思想,根据左右端点选择插入/删除直线,插入的时候利用权值树状数组维护其\(y_i\)值。之后对于竖直的两条直线,查询范围在\(low\)~\(high\)之间的水平直线个数即可。设其为\(x\),那么对答案的贡献就为\(C_{x}^2\)
注意一下\(low\)可能大于\(high\)的情况。
代码如下:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 50005;
int n;
int x[N][2], y[N][2];
int X[N], Y[N];
struct seg1{
    int x, y1, y2;
    bool operator < (const seg1 &A) const {
        return x < A.x;
    }
}s1[N];
struct seg2{
    int y, x1, x2;
}s2[N];
vector <int> v1[N], v2[N], v3[N];
int c[N];
int lowbit(int x) {
    return x & (-x);
}
void add(int x, int v) {
    for(int i = x; i < N; i += lowbit(i)) c[i] += v;
}
int query(int p) {
    int ans = 0;
    for(int i = p; i; i -= lowbit(i)) ans += c[i];
    return ans;
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin >> n;
    int D1 = 0, D2 = 0;
    for(int i = 1; i <= n; i++) {
        cin >> x[i][0] >> y[i][0] >> x[i][1] >> y[i][1];
        X[++D1] = x[i][0]; X[++D1] = x[i][1];
        Y[++D2] = y[i][0]; Y[++D2] = y[i][1];
    }
    sort(X + 1, X + D1 + 1); sort(Y + 1, Y + D2 + 1);
    D1 = unique(X + 1, X + D1 + 1) - X - 1;
    D2 = unique(Y + 1, Y + D2 + 1) - Y - 1;
    int t1 = 0, t2 = 0;
    for(int i = 1; i <= n; i++) {
        x[i][0] = lower_bound(X + 1, X + D1 + 1, x[i][0]) - X;
        x[i][1] = lower_bound(X + 1, X + D1 + 1, x[i][1]) - X;
        y[i][0] = lower_bound(Y + 1, Y + D2 + 1, y[i][0]) - Y;
        y[i][1] = lower_bound(Y + 1, Y + D2 + 1, y[i][1]) - Y;
        if(x[i][0] > x[i][1]) swap(x[i][0], x[i][1]);
        if(y[i][0] > y[i][1]) swap(y[i][0], y[i][1]);
        if(x[i][0] == x[i][1]) s1[++t1] = seg1{x[i][0], y[i][0], y[i][1]};
        if(y[i][0] == y[i][1]) s2[++t2] = seg2{y[i][0], x[i][0], x[i][1]};
        if(y[i][0] == y[i][1]) v1[x[i][0]].push_back(t2);
    }
    sort(s1 + 1, s1 + t1 + 1);
    for(int i = 1; i <= t1; i++) v3[s1[i].x].push_back(i);
    ll res = 0;
    int bound = D1;
    for(int x = 1; x <= bound; x++) {
        for(auto it : v2[x]) {
            add(s2[it].y, -1);
        }
        for(auto it : v1[x]) {
            add(s2[it].y, 1);
            v2[s2[it].x2].push_back(it);
        }
        for(auto i : v3[x]) {
            for(int j = x + 1; j <= bound; j++) {
                for(auto k : v3[j]) {
                    int R = min(s1[i].y2, s1[k].y2), L = max(s1[i].y1, s1[k].y1);
                    if(L > R) continue;
                    int tmp = query(R) - query(L - 1);
                    res += 1ll * tmp * (tmp - 1) / 2;
                }
                for(auto it : v2[j]) add(s2[it].y, -1);
            }
            for(int j = x + 1; j <= bound; j++) {
                for(auto it : v2[j]) add(s2[it].y, 1);
            }
        }
    }
    cout << res;
    return 0;
}

posted @ 2019-08-04 19:49  heyuhhh  阅读(218)  评论(0编辑  收藏  举报