BZOJ 1935: [Shoi2007]Tree 园丁的烦恼 +CDQ分治

1935: [Shoi2007]Tree 园丁的烦恼

参考与学习：https://www.cnblogs.com/mlystdcall/p/6219421.html

题意

　　在一个二维平面中有n颗树，有m次询问，要求回答在一个矩形方框中的树的个数。

思路

　　这是一个（x,y)为偏序的题目。这道题先用CDQ对x进行排序降维。然后利用树状数组对y进行处理。复杂度为O（N*logN * logN）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B; A <= C; ++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

/*-----------------------show time----------------------*/
            const int maxl = 10000009;
            const int maxn = 500009;

            struct node
            {
                int id,x,y,type,val;

                bool operator<(const node & a)const{
                    if(a.x == x)return type < a.type;
                    return x < a.x;             //按照x排序
                }                
            }a[maxn*5],b[maxn*5];
            ll ans[maxn];
            int n,m,tot,askid,maxy = -1;

            void add(int type,int x,int y,int val,int id){
                tot++;
                a[tot].type = type;
                a[tot].x = x;
                a[tot].y = y;
                a[tot].val = val;
                a[tot].id = id;
            }
            ll sum[maxl];       //树状数组
            int lowbit(int x){
                return x & (-x);
            }
            void update(int x,int c){
                while(x  <= maxy){
                    sum[x]+=c;
                    x += lowbit(x);
                }
            }
            ll query(int x){
                ll cnt = 0;
                while(x > 0){
                    cnt += sum[x];
                    x -= lowbit(x);
                }
                return cnt;
            }
            void clearb(int x){
                while(x <= maxy){
                    if(sum[x])sum[x] = 0;
                    else break;
                    x += lowbit(x);
                }
            }
            void CDQ(int L, int R){
                if(L >= R)return ;
                int mid = (L + R)>>1;
                CDQ(L, mid); CDQ(mid+1, R);

                int q1 = L, q2 = mid+1;
                for(int i=L; i<=R; i++){
                    if((q1 <= mid && a[q1] < a[q2]) || q2 > R){
                        if(a[q1].type == 0){
                            update(a[q1].y, 1);
                        }
                        b[i] = a[q1++];
                    }
                    else {
                        if(a[q2].type == 1)ans[a[q2].id] += a[q2].val * query(a[q2].y);
                        b[i] = a[q2++];
                    }
                }
                for(int i=L; i<=R;i++){
                    a[i] = b[i];
                    clearb(b[i].y);
                }

            }
int main(){
            scanf("%d%d", &n, &m);
            for(int i=1; i<=n; i++){

                int x,y;
                tot++;
                scanf("%d%d", &x, &y);x+=2,y+=2;
                a[tot].x = x;a[tot].y = y;
                a[tot].id = 0;a[tot].type = 0;
                maxy = max(maxy, y);
            }

            for(int i=1; i<=m; i++){
                int x1,y1,x2,y2;
                askid++;
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                x1+=2,y1+=2,x2+=2,y2+=2;
                maxy = max(maxy, max(y1,y2));
                add(1,x1-1,y1-1,1,askid);
                add(1,x1-1,y2,-1,askid);
                add(1,x2,y1-1,-1,askid);
                add(1,x2,y2,1,askid);
            }   
            CDQ(1,tot);
            for(int i=1; i<=askid; i++){
                printf("%lld
", ans[i]);
            }
            return 0;
}