POJ 2155 Matrix

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 24547		Accepted: 9085

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

 

 

 

解析：二维树状数组（区间更新，单点查询）。解答原理参考《浅谈信息学竞赛中的“0”和“1”》。

 

 

 

#include <cstdio>
#include <cstring>
#define lowbit(x) (x)&(-x)

int n, t, q;
int c[1005][1005];
char op[2];
int x1, y1, x2, y2;

void add(int x, int y, int val)
{
    for(int i = x; i <= n; i += lowbit(i))
        for(int j = y; j <= n; j += lowbit(j))
            c[i][j] += val;
}

int sum(int x, int y)
{
    int ret = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        for(int j = y; j > 0; j -= lowbit(j))
            ret += c[i][j];
    return ret;
}

void solve()
{
    scanf("%d%d", &n, &q);
    memset(c, 0, sizeof(c));
    while(q--){
        scanf("%s", op);
        if(op[0] == 'C'){
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            add(x1, y1, 1);
            add(x2+1, y2+1, 1);
            add(x1, y2+1, 1);
            add(x2+1, y1, 1);
        }
        else{
            scanf("%d%d", &x1, &y1);
            printf("%d
", sum(x1, y1)&1);
        }
    }
}

int main()
{
    scanf("%d", &t);
    while(t--){
        solve();
        printf("
");
    }
    return 0;
}