POJ 2155 Matrix 【二维线段树模板题】

Description

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

只不过是结构体当中又套了一个而已，模板题不解释

#include <stdio.h>
#include <string.h>
#include <iostream>

using namespace std;
const int MAXN = 1010;
struct Nodey
{
int l,r;
int val;
};
int n;
int locx[MAXN],locy[MAXN];
struct Nodex
{
int l,r;
Nodey sty[MAXN*3];
void build(int i,int _l,int _r)
{
sty[i].l = _l;
sty[i].r = _r;
sty[i].val = 0;
if(_l == _r)
{
locy[_l] = i;
return;
}
int mid = (_l + _r)>>1;
build(i<<1,_l,mid);
build((i<<1)|1,mid+1,_r);
}
void add(int i,int _l,int _r,int val)
{
if(sty[i].l == _l && sty[i].r == _r)
{
sty[i].val += val;
return;
}
int mid = (sty[i].l + sty[i].r)>>1;
if(_r <= mid)add(i<<1,_l,_r,val);
else if(_l > mid)add((i<<1)|1,_l,_r,val);
else
{
add(i<<1,_l,mid,val);
add((i<<1)|1,mid+1,_r,val);
}
}
}stx[MAXN*3];
void build(int i,int l,int r)
{
stx[i].l = l;
stx[i].r = r;
stx[i].build(1,1,n);
if(l == r)
{
locx[l] = i;
return;
}
int mid = (l+r)>>1;
build(i<<1,l,mid);
build((i<<1)|1,mid+1,r);
}
void add(int i,int x1,int x2,int y1,int y2,int val)
{
if(stx[i].l == x1 && stx[i].r == x2)
{
stx[i].add(1,y1,y2,val);
return;
}
int mid = (stx[i].l + stx[i].r)/2;
if(x2 <= mid)add(i<<1,x1,x2,y1,y2,val);
else if(x1 > mid)add((i<<1)|1,x1,x2,y1,y2,val);
else
{
add(i<<1,x1,mid,y1,y2,val);
add((i<<1)|1,mid+1,x2,y1,y2,val);
}
}
int sum(int x,int y)
{
int ret = 0;
for(int i = locx[x];i;i >>= 1)
for(int j = locy[y];j;j >>= 1)
ret += stx[i].sty[j].val;
return ret;
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
scanf("%d",&T);
while(T--)
{
int q;
scanf("%d%d",&n,&q);
build(1,1,n);
char op[10];
int x1,x2,y1,y2;
while(q--)
{
scanf("%s",op);
if(op[0] == 'C')
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
add(1,x1,x2,y1,y2,1);
}
else
{
scanf("%d%d",&x1,&y1);
if(sum(x1,y1)%2 == 0)printf("0\n");
else printf("1\n");
}
}
if(T)printf("\n");
}
return 0;
}