Matrix

                                                                                                                                    Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 25368 Accepted: 9403

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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

题意：给一个N*N的矩阵，里面的值不是0，就是1。初始时每一个格子的值为0。 现对该矩阵有两种操作：（共T次） 1.C x1 y1 x2 y2：将左上角为(x1, y1)，右下角为(x2, y2)这个范围的子矩阵里的值全部取反。 2.Q x y：查询矩阵中第i行，第j列的值。 (2 <= N <= 1000, 1 <= T <= 50000) 思路：参见国家集训队论文：武森《浅谈信息学竞赛中的“0”和“1”》 1. 根据这个题目中介绍的这个矩阵中的数的特点不是 1 就是 0，这样我们只需记录每个格子改变过几次，即可判断这个格子的数字。 2. 先考虑一维的情况： 若要修改[x,y]区间的值，其实可以先只修改 x 和 y+1 这两个点的值（将这两个点的值加1）。查询k点的值时，其修改次数即为 sum(cnt[1] + … + cnt[k])。 3. 二维的情况： 道理同一维。要修改范围[x1, y1, x2, y2]，只需修改这四个点：(x1,y1), (x1,y2+1), (x2+1,y1), (x2+1,y2+1)。查询点(x,y)的值时，其修改次数为 sum(cnt[1, 1, x, y])。 4. 而区间求和，便可用树状数组来实现。

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int c[2000][2000];int n,m;int lowbit(int i){    return i&-i;}void add(int i,int j,int zhi){    for(int k=i; k<2020; k+=lowbit(k))    {        for(int p=j; p<2020; p+=lowbit(p))        {            c[k][p]+=zhi;        }    }}int sum(int i,int j){    int he=0;    for(int k=i; k>0; k-=lowbit(k))    {        for(int p=j; p>0; p-=lowbit(p))        {            he+=c[k][p];        }    }    return he;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&n,&m);        memset(c,0,sizeof(c));        while(m--)        {            char a[10];            int x,y;            scanf("%s%d%d",a,&x,&y);            if(a[0]=='C')            {                int ha,haha;                scanf("%d%d",&ha,&haha);                add(x, y, 1);                add(x, haha+1, 1);                add(ha+1, y, 1);                add(ha+1, haha+1, 1);            }            else            {                printf("%d\n",sum(x,y)%2);            }        }        printf("\n");    }}