bzoj4170 极光

题目链接

题面

题意

把每个位置的点都看成是一个二维坐标系中的点。比如第\(i\)个点就是\((i,a[i])\)。

有两种操作

询问:然后每次询问的就是与当前点坐标的曼哈顿距离小于等于\(k\)的点。

修改:修改第i个点的纵坐标。保留历史点。

思路

旋转坐标系。然后就变成了添加一个点和询问一个子矩阵内点的个数。用\(CDQ\)分治或者其他数据结构做就行了

代码

/*
* @Author: wxyww
* @Date:   2019-02-15 08:38:47
* @Last Modified time: 2019-02-15 10:16:11
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 4000010,M = 3000003,BL = 1000000;
#define  pi pair<int,int>
ll read() {
    ll x=0,f=1;char c=getchar();
    while(c<'0'||c>'9') {
        if(c=='-') f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9') {
        x=x*10+c-'0';
        c=getchar();
    }
    return x*f;
}
struct node {
    int x,y,opt,pos,val;
}a[M];
char s[10];
int tt[N],tot;
int tree[M],mx,ans[N];
vector<pi>v;
void update(int pos,int c) {
    while(pos <= BL*3) {
        tree[pos] += c;
        pos += pos & -pos;
    }
}
int query(int pos) {
    int ret = 0;
    while(pos) {
        ret += tree[pos];
        pos -= pos & -pos;
    }
    return ret;
}
node tmp[N];
void cdq(int l,int r) {
    if(r <= l) return;
    int mid = (l + r) >> 1;
    cdq(l,mid);cdq(mid + 1,r);
    int now = l,L = l,R = mid + 1;
    while(L <= mid && R <= r) {
        if(a[L].x <= a[R].x) {
            if(a[L].opt == 1) {
                update(a[L].y,a[L].val);
                v.push_back(make_pair(a[L].y,a[L].val));
            }
            tmp[now++] = a[L++];
        }
        else {
            if(a[R].opt == 2) ans[a[R].pos] += query(a[R].y);
            else if(a[R].opt == 3) ans[a[R].pos] -= query(a[R].y);
            tmp[now++] = a[R++];
        }
    }
    while(L <= mid) tmp[now++] = a[L++];
    while(R <= r) {
        if(a[R].opt == 2) ans[a[R].pos] += query(a[R].y);
        else if(a[R].opt == 3) ans[a[R].pos] -= query(a[R].y);
        tmp[now++] = a[R++];
    }
    for(int i = l;i <= r;++i) a[i] = tmp[i];
    int k = v.size();
    for(int i = 0;i < k;++i) update(v[i].first,-v[i].second);
    v.clear();
}
int main() {
    int n = read(),Q = read(),now = 0;
    for(int i = 1;i <= n;++i) {
        tt[i] = read();
        a[++tot].x = i + tt[i],a[tot].y = i - tt[i] + BL;
        a[tot].opt = 1;a[tot].val = 1;
        mx = max(mx,max(a[tot].x,a[tot].y));
    }
    for(int i = 1;i <= Q;++i) {
        scanf("%s",s + 1);
        int x = read(),y = read();
        if(s[1] == 'Q') {
            int x1 = x + tt[x] - y,x2 = x + tt[x] + y,y1 = x - tt[x] + BL - y,y2 = x - tt[x] + BL + y;
            a[++tot].pos = ++now;a[tot].x = x2;a[tot].y = y2;a[tot].opt = 2;
            a[++tot].pos = now;a[tot].x = x1 - 1;a[tot].y = y1 - 1;a[tot].opt = 2;
            a[++tot].pos = now;a[tot].x = x1 - 1;a[tot].y = y2;a[tot].opt = 3;
            a[++tot].pos = now;a[tot].x = x2;a[tot].y = y1 - 1;a[tot].opt = 3;
        }
        else {tt[x] = y;
            a[++tot].opt = 1;
            a[tot].x = x + tt[x];a[tot].y = x - tt[x] + BL;a[tot].val = 1;
            mx = max(mx,max(a[tot].x,a[tot].y));
        }
    }    cdq(1,tot);
    for(int i = 1;i <= now;++i) printf("%d\n",ans[i]);
    return 0;
}