POJ 2155 Matrix 二维树状数组

Matrix
Time Limit: 3000MS      Memory Limit: 65536K
Total Submissions: 25586        Accepted: 9470
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
Source

POJ Monthly,Lou Tiancheng

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>

const int N = 1000+5;

struct BIT{
int n;
int c

;
void add(int x,int y,int num){
for(int i=x;i<=n;i+=i&(-i)){
for(int j=y;j<=n;j+=j&(-j)){
c[i][j]+=num;
}
}
}
int sum(int x,int y){
int ans=0;
for(int i=x;i;i-=i&(-i)){
for(int j=y;j;j-=j&(-j)){
ans+=c[i][j];
}
}
return ans;
}
void resize(int n){
this->n=n;
for(int i=1;i<=n;++i){
fill(c[i],c[i]+n+1,0);
}
}
}bit;

void operC(int x1,int y1,int x2,int y2){
bit.add(x1,y1,1);
bit.add(x1,y2+1,-1);
bit.add(x2+1,y1,-1);
bit.add(x2+1,y2+1,1);//此处被+1 再2次被-1 加1补回0
}

inline int operQ(int x,int y){
return bit.sum(x,y);
}

int main()
{
//freopen("/home/lu/文档/r.txt","r",stdin);
//freopen("/home/lu/文档/w.txt","w",stdout);
int X,n,t;
char ch;
int x1,x2,y1,y2;
scanf("%d",&X);
while(X--){
scanf("%d%d",&n,&t);
bit.resize(n);
while(t--){
scanf(" %c",&ch);
if(ch=='C'){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
operC(x1,y1,x2,y2);
}
else{
scanf("%d%d",&x1,&y1);
printf("%d\n",operQ(x1,y1)&1);
}
}
if(X!=0){
putchar('\n');
}
}
return 0;
}