題目連結【簡單題】

  一道簡單的二維線段樹，查詢的是一個規定長寬的長方形內的最多點數，因為W、H都比較的小，所以不妨可以直接上暴力二維線段樹查詢即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define efs 1e-6
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 105;
int a[maxN][maxN], W, H, N, S, T, tree[maxN<<2][maxN<<2];
void build_In(int rt, int xo, int pos, int l, int r, bool flag)
{
    if(l == r)
    {
        if(flag) tree[xo][rt] = a[pos][l];
        else tree[xo][rt] = tree[xo<<1][rt] + tree[xo<<1|1][rt];
        return;
    }
    int mid = (l + r)>>1;
    build_In(rt<<1, xo, pos, l, mid, flag);
    build_In(rt<<1|1, xo, pos, mid+1, r, flag);
    tree[xo][rt] = tree[xo][rt<<1] + tree[xo][rt<<1|1];
}
void build_Out(int rt, int l, int r)
{
    if(l == r)
    {
        build_In(1, rt, l, 1, H, true);
        return;
    }
    int mid = (l + r)>>1;
    build_Out(rt<<1, l, mid);
    build_Out(rt<<1|1, mid+1, r);
    build_In(1, rt, 0, 1, H, false);
}
int query_In(int rt, int xo, int l, int r, int ql, int qr)
{
    if(ql<=l && qr>=r) return tree[xo][rt];
    int mid = (l + r)>>1;
    if(ql>mid) return query_In(rt<<1|1, xo, mid+1, r, ql, qr);
    else if(qr<=mid) return query_In(rt<<1, xo, l, mid, ql, qr);
    else
    {
        int ans = query_In(rt<<1|1, xo, mid+1, r, ql, qr);
        ans += query_In(rt<<1, xo, l, mid, ql, qr);
        return ans;
    }
}
int query_Out(int rt, int l, int r, int qlx, int qly, int qrx, int qry)
{
    if(qlx<=l && qrx>=r) return query_In(1, rt, 1, H, qly, qry);
    int mid = (l + r)>>1;
    if(qlx>mid) return query_Out(rt<<1|1, mid+1, r, qlx, qly, qrx, qry);
    else if(qrx<=mid) return query_Out(rt<<1, l, mid, qlx, qly, qrx, qry);
    else
    {
        int ans = query_Out(rt<<1|1, mid+1, r, qlx, qly, qrx, qry);
        ans += query_Out(rt<<1, l, mid, qlx, qly, qrx, qry);
        return ans;
    }
}
int main()
{
    while(scanf("%d", &N) && N)
    {
        scanf("%d%d", &W, &H);
        memset(a, 0, sizeof(a));
        for(int i=1; i<=N; i++)
        {
            int e1, e2;
            scanf("%d%d", &e1, &e2);
            a[e1][e2]++;
        }
        build_Out(1, 1, W);
        scanf("%d%d", &S, &T);
        int ans = 0;
        for(int i=1; i+S-1<=W; i++)
        {
            for(int j=1; j+T-1<=H; j++)
            {
                ans = max(ans, query_Out(1, 1, W, i, j, i+S-1, j+T-1));
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}