题目链接

题意：给一个平面，有两个操作

• 0 x y 在(x,y)处放置一个点
• 1 x1 y1 x2 y2 查询左下角为(x1,y1)，右上角为(x2,y2)的矩形区域包含多少个点（包括边界）

使用二维树状数组，矩形面积相减得到目标区域的和
二维数组模板题

"math.h" 中包含了对变量 x1,y1 的定义

题意中的位置是0~1000

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <string>
//#include <math.h>
#include <bitset>
#include <ctype.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
//const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 1200 + 5;
const int mod = 1e9 + 7;
int t, kase = 0;
int c[N][N];
int n;
void init()
{memset(c, 0, sizeof(c));
}
inline int lowbit(int i)
{return i&(-i);
}
void add(int x, int y, int val)
{for(int i = x; i <= n; i += lowbit(i)){for(int j = y; j <= n; j += lowbit(j)){c[i][j] += val;}}
}
int sum(int x, int y)
{int ans = 0;for(int i = x; i > 0; i -= lowbit(i))for(int j = y; j > 0; j -= lowbit(j))ans += c[i][j];return ans;
}
int query(int x1, int y1, int x2, int y2)
{return sum(x2,y2) - sum(x2, y1-1) - sum(x1-1, y2) + sum(x1-1, y1-1);
}
int x1,x2,y1,y2,q,type;
int main()
{n = 1001;scanf("%d", &t);while(t--){memset(c, 0, sizeof(c));printf("Case %d:\n", ++kase);scanf("%d", &q);while(q--){scanf("%d", &type);if(type == 0){scanf("%d%d", &x1,&y1);x1++,y1++;if(!query(x1,y1,x1,y1))add(x1,y1,1);}else{scanf("%d%d%d%d", &x1, &y1, &x2, &y2);x1++,y1++,x2++,y2++;printf("%d\n", query(x1,y1,x2,y2));}}}return 0;
}