Matrix
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 21757 | Accepted: 8141 |
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].
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.
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.
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
题意是起始给出一个全为0的矩阵,然后不断地对其子矩阵操作,0变为1,1变为0。然后也不断查询某一个位置的值。
二维树状数组,之前一直理解错了,其实转换的那四块是对上面的部分,不是下面的部分。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std; #define MY_MAX 1100 int tree[MY_MAX*3][MY_MAX*3]; int n,t; int lowbit(int x) { return x&(-x); } int get_num(int x,int y) { int sum=0,i,j; for(i=x;i>0;i=i-lowbit(i)) { for(j=y;j>0;j=j-lowbit(j)) { sum += tree[i][j]; } } return sum; } void cal(int x,int y) { int i,j; for(i=x;i<=n;i=i+lowbit(i)) { for(j=y;j<=n;j=j+lowbit(j)) { tree[i][j]++; } } } int main() { //freopen("i.txt","r",stdin); //freopen("o.txt","w",stdout); int test,i,x1,y1,x2,y2,x,y; char oper[10]; scanf("%d",&test); while(test--) { scanf("%d%d",&n,&t); memset(tree,0,sizeof(tree)); for(i=1;i<=t;i++) { scanf("%s",oper); if(oper[0]=='C') { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); cal(x2+1,y2+1); cal(x1,y1); cal(x1,y2+1); cal(x2+1,y1); } else if(oper[0]=='Q') { scanf("%d%d",&x,&y); printf("%d ",get_num(x,y)%2); } } printf(" "); } //system("pause"); return 0; }
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