POJ-2155：Matrix（二维树状数祖）

Posted 2020-11-17 goldenfingers

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 31892		Accepted: 11594

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

解体心得：

• 题意就是给你一个矩阵，你可以选择一个子矩阵来进行0-1反转，然后单点查询0-1
• 刚开始什么都没想，直接写了一个二维线段树，非常暴力，复杂度是n²logn，然后理所当然的TLE。
• 后来发现线段树就维护一个0-1值十分的浪费，然后想了一下不就是一个树状数组就可以解决。直接记录变换次数就行了，偶数次变换就是没有变化，然后记录前缀和。但是这个题要注意的是用二维树状数组。二维树状数组的用法直接看代码就行了。
• 在写二维树状数组的时候要注意就是前缀和记录时要容斥定理解决重复区间维护的问题。

 

#include <algorithm>
#include <stdio.h>
#include <cstring>
using namespace std;
const int maxn = 1010;

int sum[maxn][maxn],n,m;

int lowbit(int x) {
    return x&-x;
}

void add(int x,int y,int va) {
    if(x < 1 || y < 1 || x > n || y > n)
        return ;
   for(int i=x;i<=n;i+=lowbit(i))
       for(int j=y;j<=n;j+=lowbit(j))
           sum[i][j] += va;
}

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

int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&n,&m);
        memset(sum,0,sizeof(sum));
        while(m--) {
            char s[3];
            scanf("%s",s);
            if(s[0] == ‘C‘) {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&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 {
                int x,y;
                scanf("%d%d",&x,&y);
                int ans = Query(x,y);
                printf("%d
",ans%2);
            }
        }
        printf("
");
    }
    return 0;
}