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POJ_2155_matrix

Posted on In

Language:
Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 37057 Accepted: 13286
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].

输入描述:

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.

输出描述:

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.

示例1
输入

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

输出

1
0
0
1

poj_2155.cppview raw
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#include <iostream>
#include <cstring>
using namespace std;

const int N = 1010;
int sum[N][N];
int n;
int C(int x){
    return x & -x;
}

int query(int x, int y){
    int ans = 0;
    while(x>0){
        int ty = y;
        while(ty>0){
            ans += sum[x][ty];
            ty -= C(ty);
        }
        x -= C(x);
    }
    return ans;
}

void add(int x, int y, int d){
    while(x<=n){
        int ty = y;
        while(ty<=n){
            sum[x][ty] = (sum[x][ty] + d) % 2;
            ty += C(ty);
        }
        x += C(x);
    }
}

int main(){
    int X ;
    scanf("%d", &X);
    int T;
    char c[10];
    int x1, y1, x2, y2;
    int x, y;
    for(int i=0;i<X;i++){
        scanf("%d%d", &n, &T);
        memset(sum, 0, sizeof sum);
        while(T--){
            scanf("%s",c);
            if(c[0]=='C'){
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
//                printf("%d %d %d %d\n", x1,y1, x2,y2);
                add(x1, y1, 1);
                add(x1, y2+1, 1);
                add(x2+1, y1, 1);
                add(x2+1, y2+1, +1);
            }else if(c[0]=='Q'){
                scanf("%d%d", &x, &y);
//                printf("%d %d\n", x, y);
                printf("%d\n", query(x, y) % 2);
            }
        }
        puts("");
    }
    return 0;
}