二维线段树
用于矩阵的快速修改和查询（修改和查询的时间复杂度都为logn，空间复杂度是（4*n）*（4*n））
这里提供两种操作 单点修改+区间询查 区间修改+单点询查
由于（单点修改+区间询查 ）和（ 区间修改+单点询查）的sum数组意义不同，所以该代码无法实现（单点修改+单点询查）和（区间修改+区间询查）
单点修改
void changex(int kx,int l,int r)
{
changey(kx,1,1,h);
if(l==r) return;
int mid=l+r>>1;
if(x<=mid) changex(kx<<1,l,mid);
else changex(kx<<1|1,mid+1,r);
}
void changey(int kx,int ky,int l,int r)
{
sum[kx][ky]++;
if(l==r) return;
int mid=l+r>>1;
if(y<=mid) changey(kx,ky<<1,l,mid);
else changey(kx,ky<<1|1,mid+1,r);
}
区间询查
void queryx(int kx,int l,int r)
{
if(l>=xl && r<=xr)
{
queryy(kx,1,1,h);
return;
}
int mid=l+r>>1;
if(xl<=mid) queryx(kx<<1,l,mid);
if(xr>mid) queryx(kx<<1|1,mid+1,r);
}
void queryy(int kx,int ky,int l,int r)
{
if(l>=yl && r<=yr)
{
cnt+=sum[kx][ky];
return;
}
int mid=l+r>>1;
if(yl<=mid) queryy(kx,ky<<1,l,mid);
if(yr>mid) queryy(kx,ky<<1|1,mid+1,r);
}
区间修改
void changex(int kx,int l,int r)
{
if(x1<=l&&r<=x2)
{
changey(kx,1,1,n);
return;
}
int mid=l+r>>1;
if(x1<=mid) changex(kx<<1,l,mid);
if(x2>mid) changex(kx<<1|1,mid+1,r);
}
void changey(int kx,int ky,int l,int r)
{
if(y1<=l&&r<=y2)
{
sum[kx][ky]++;
return;
}
int mid=l+r>>1;
if(y1<=mid) changey(kx,ky<<1,l,mid);
if(y2>mid) changey(kx,ky<<1|1,mid+1,r);
}
单点询查
void queryx(int kx,int l,int r)
{
queryy(kx,1,1,n);
if(l==r) return;
int mid=l+r>>1;
if(x<=mid) queryx(kx<<1,l,mid);
else queryx(kx<<1|1,mid+1,r);
}
void queryy(int kx,int ky,int l,int r)
{
ans+=sum[kx][ky];
if(l==r) return;
int mid=ly+ry>>1;
if(y<=mid) queryy(kx,ky<<1,l,mid);
else queryy(kx,ky<<1|1,mid+1,r);
}
Matrix POJ - 2155 （二维线段树区间修改+区间询查）
https://vjudge.net/contest/301590#problem/D
Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).
We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].
Input
The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.
Output
For each querying output one line, which has an integer representing A[x, y].
There is a blank line between every two continuous test cases.
Sample Input
1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1
Sample Output
1
0
0
1#include
#include
const int num=1005;
int tree[num*4][num*4];
int x1,y1,x2,y2,n,ans;
int pointx,pointy;
void changey(int kx,int ky,int treel,int treer)
{
if(y1<=treel&&y2>=treer){
tree[kx][ky]++;
return;
}
int bet=(treel+treer)/2;
if(y1<=bet) changey(kx,ky*2,treel,bet);
if(y2>bet) changey(kx,ky*2+1,bet+1,treer);
}
void changex(int kx,int treel,int treer)
{
if(x1<=treel&&x2>=treer){
changey(kx,1,1,n);
return;
}
int bet=(treel+treer)/2;
if(x1<=bet) changex(kx*2,treel,bet);
if(x2>bet) changex(kx*2+1,bet+1,treer);
}
void asky(int kx,int ky,int treel,int treer)
{
ans+=tree[kx][ky];
if(treel==treer)
return;
int bet=(treel+treer)/2;
if(pointy<=bet) asky(kx,ky*2,treel,bet);
else asky(kx,ky*2+1,bet+1,treer);
}
void askx(int kx,int treel,int treer)
{
asky(kx,1,1,n);
if(treel==treer)
return;
int bet=(treel+treer)/2;
if(pointx<=bet) askx(kx*2,treel,bet);
else askx(kx*2+1,bet+1,treer);
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
memset(tree,0,sizeof(tree));
int m;
scanf("%d %d",&n,&m);
for(int i=0;i