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Matrix
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 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 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 |
题意：给出矩阵左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1，然后给出询问，问某个点是多少。
题解：纠结了好久，看了这篇博客后秒懂http://blog.sina.com.cn/s/blog_626489680100k75p.html
先举个一维的样例：你要使区间[x,y]所有加上一个值v,结合树状数组的功能，能够类似扫气球那样，在x处加v, y+1处减1
这样假设你要求x处的值，就转换成求[1,x]的和了，比如 ：一个n=6的数组，一開始为0 0 0 0 0 0
在[2,4]加上2后变成 0 2 0 0 -2 0 这样前缀和 sum[1]=0;sum[2]=2;sum[3]=2;sum[4]=2;sum[5]=0;sum[6]=0;
依次代表了每一个数的值。
二维的也一样，由于二维树状数组的getsum(int x,int y)函数是求矩阵（1,1）~(x,y)的值得和。也就类似于前缀和。原理和一维的一样
仅仅只是线操作改成了平面操作。自己能够画个图感受下。
即：
add(x,y,1);
add(x,y1+1,-1);
add(x1+1,y,-1);
add(x1+1,y1+1,1);
然后查询单点就是求和了。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#define N 1040
#define ll long long
using namespace std;
int n;
int bit[N][N];
int sum(int i,int j) {
int s=0;
while(i>0) {
int jj=j;
while(jj>0) {
s+=bit[i][jj];
jj-=jj&-jj;
}
i-=i&-i;
}
return s;
}
void add(int i,int j,int x) {
while(i<=n) {
int jj=j;
while(jj<=n) {
bit[i][jj]+=x;
jj+=jj&-jj;
}
i+=i&-i;
}
}
int main() {
freopen("test.in","r",stdin);
int t;
cin>>t;
while(t--) {
int q;
scanf("%d%d ",&n,&q);
memset(bit,0,sizeof bit);
char c;
int x,y,x1,y1;
while(q--) {
scanf("%c",&c);
if(c==‘C‘) {
scanf("%d%d%d%d",&x,&y,&x1,&y1);
add(x,y,1);
add(x,y1+1,-1);
add(x1+1,y,-1);
add(x1+1,y1+1,1);//重叠的部分加上
} else {
scanf("%d%d",&x,&y);
printf("%d\n",sum(x,y)%2);
}
getchar();
}
if(t)printf("\n");
}
return 0;
}