目录

• 题目链接
• 题解
• 代码

题目链接

bzoj4237: 稻草人

题解

暴力统计是n^2的
考虑统计一段区间对另一端的贡献
对于y值cdq分治，降调一维
对于当前两个分治区间统计上面那部分对下面那部分的贡献
对当前两区间x排序后，对上部分维护单增单调栈，得到距离当前点最近的比她低的点p
对于下面的区间维护一个上凸壳 ，直接在凸壳上二分p统计答案

代码

#include<set>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar()
#define pc putchar
#define LL long long
inline int read() {
    int x = 0,f = 1;
    char c = gc;
    while(c < '0' || c > '9') c = gc;
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
    return x * f;
}
void print(LL x) {
    if(x < 0) {
        pc('-');
        x = -x;
    }
    if(x >= 10) print(x / 10);
    pc(x % 10 + '0');
}
const int maxn = 200007;
struct Point {
    int x,y;
} po[maxn];
int n;
bool cmpx(Point a,Point b) { return a.x < b.x;  }
bool cmpy(Point a,Point b) { return a.y < b.y;  }
int tp,tl;
int sk[maxn],sk2[maxn];
LL  ans = 0;
void solve(int l = 1,int r = n) {
    if(l == r) return ;
    int mid = l + r >> 1;
    std::sort(po + l,po + r + 1,cmpy);
    std::sort(po + l,po + mid + 1,cmpx); //down
    std::sort(po + mid + 1,po + r + 1,cmpx); //up
    tp = tl = 0;
    int p = l,L,R,to,miid,lim;
    for(int i = mid + 1;i <= r;++ i) {
        while(tp && po[sk[tp]].y >= po[i].y) tp --;
        sk[++ tp] = i; 

        for(;p <= mid && po[p].x < po[i].x;++ p) {
            while(tl && po[sk2[tl]].y <= po[p].y) tl --;
            sk2[++ tl] = p;
        } 

        L = 1,R = tl;to = - 1;lim = po[sk[tp - 1]].x;
        while(L <= R) {
            miid = L + R >> 1;
            if(po[sk2[miid]].x > lim) to = miid,R = miid - 1;
            else L = miid + 1;
        }
        if(to != -1) ans += 1ll * tl - 1ll * to + 1;
    }
    solve(l,mid); solve(mid + 1,r);
}
int main() {
    n = read();
    for(int i = 1;i <= n;++ i) {
        po[i].x = read(),po[i].y = read();
    }
    po[0].x = po[0].y = -1;
    solve();
    print(ans);
    pc('\n');
} 

原文地址：https://www.cnblogs.com/sssy/p/9715479.html