Bzoj4237：稻草人

题面

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Sol

\(CDQ\)分治

先对\(x\)排序，对\(y\)在\(CDQ\)分治是从大到小排序

从大到小加入，右边用单调栈维护\(x\)递增，\(y\)递减的序列

左边就是找到\(x\)比它大，最小的\(y\)（树状数组解决）

再在右边找到最后一个小于这个\(y\)的位置，那么栈顶到这个位置都是答案

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e5 + 5);

IL ll Input(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, bit[_];
ll ans;
struct Data{
    int x, y, id;
    IL bool operator <(RG Data B) const{
        return x < B.x;
    }
} q[_], p[_], S[_];

IL void Modify(RG int x, RG int d){
    for(; x <= n; x += x & -x) bit[x] = min(bit[x], d);
}

IL void Clear(RG int x){
    for(; x <= n; x += x & -x) bit[x] = bit[0];
}

IL int Query(RG int x){
    RG int ret = bit[0];
    for(; x; x -= x & -x) ret = min(ret, bit[x]);
    return ret;
}

IL void CDQ(RG int l, RG int r){
    if(l == r) return;
    RG int mid = (l + r) >> 1;
    CDQ(l, mid); CDQ(mid + 1, r);
    for(RG int i = l, j = mid + 1, k = l; k <= r; ++k)
        if(j > r || (i <= mid && p[i].y > p[j].y)) q[k] = p[i++];
        else q[k] = p[j++];
    for(RG int i = l; i <= r; ++i) p[i] = q[i];
    RG int top = 0;
    for(RG int i = l; i <= r; ++i)
        if(p[i].id <= mid){
            RG int v = Query(n - p[i].id + 1);
            Modify(n - p[i].id + 1, p[i].y);
            RG int ll = 1, rr = top, cnt = 0;
            while(ll <= rr){
                RG int mmid = (ll + rr) >> 1;
                if(S[mmid].y < v) cnt = mmid, rr = mmid - 1;
                else ll = mmid + 1;
            }
            if(cnt) ans += top - cnt + 1;
        }
        else{
            while(top && S[top].x > p[i].x) --top;
            S[++top] = p[i];
        }
    for(RG int i = l; i <= r; ++i)
        if(p[i].id <= mid) Clear(n - p[i].id + 1);
}

int main(RG int argc, RG char* argv[]){
    n = Input(); Fill(bit, 127);
    for(RG int i = 1; i <= n; ++i) p[i].x = Input(), p[i].y = Input();
    sort(p + 1, p + n + 1);
    for(RG int i = 1; i <= n; ++i) p[i].id = i;
    CDQ(1, n);
    printf("%lld\n", ans);
    return 0;
}