目錄

• 題目鏈接
• 題解
• 代碼

題目鏈接

bzoj4237: 稻草人

題解

暴力統計是n^2的
考慮統計一段區間對另一端的貢獻
對於y值cdq分治，降調一維
對於當前兩個分治區間統計上面那部分對下面那部分的貢獻
對當前兩區間x排序後，對上部分維護單增單調棧，得到距離當前點最近的比她低的點p
對於下面的區間維護一個上凸殼 ，直接在凸殼上二分p統計答案

代碼

#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'); 
} 
 

bzoj4237: 稻草人 cdq分治 單調棧