POJ - 2155 Matrix (二维树状数组 + 区间改动 + 单点求值 或者 二维线段树 + 区间更新 + 单点求值)

Matrix

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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

题目意思是给你一个矩阵，最開始都是为0，然后给你一个左上角的坐标，右下角的坐标，将这个区域的0 -> 1,1 -> 0，然后给你一个点让你求出他此时的结果是1还是0.

这里能够用树状数组的区间改动以及单点求值。仅仅是这里是二维的树状数组，原理是一样的。

将其改变的话，能够如此，如果最開始[1....x2][1....y2]为偶数的话(这里就是看成是前辍和)

先将[1....x2][1....y2]添加1,变为奇数,然后将[1...x1 - 1][y2]添加1变为偶数,将[1....x2][y1]添加1变为偶数,而他们中间重叠的[1....x1 - 1][1....x2 - 1]部分则是被改变了三次。还是奇数，所以还有加1。将他变为偶数,如此除了[x1....x2][y1....y2]变了外。其它的地方依然没有改变。

/*
Author: 2486
Memory: 4304 KB		Time: 547 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1000 + 5;
int C[MAXN][MAXN];
int N, X, M, x, y, x1, x2, y1, y2;
char op[10];
int lowbits(int x){
    return x & (-x);
}

void add(int x,int y){
    for(int i = x; i <= N;i += lowbits(i)){
        for(int j = y;j <= N;j += lowbits(j)){
            C[i][j] ++;
        }
    }
}

int query(int x,int y){
    int ret = 0;
    for(int i = x;i > 0;i -= lowbits(i)){
        for(int j = y;j > 0;j -= lowbits(j)){
            ret += C[i][j];
        }
    }
    return ret;
}

int main(){
    //freopen("D://imput.txt","r",stdin);
    scanf("%d", &X);
    while(X --){
        memset(C, 0, sizeof(C));
        scanf("%d%d", &N, &M);
        while(M --){
            scanf("%s", op);
            if(op[0] == ‘C‘){
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                x2 ++;
                y2 ++;
                add(x1, y1);
                add(x1, y2);
                add(x2, y1);
                add(x2, y2);
            }
            else{
                scanf("%d%d", &x, &y);
                printf("%d\n", query(x,y) & 1);//求和即代表求点
            }
        }
        if(X)printf("\n");
    }
    return 0;
}

线段树代码：

/*
Author: 2486
Memory: 63688 KB		Time: 1579 MS
Language: G++		Result: Accepted
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define root 1, 1, N
const int MAXN = 1000 + 5;
int N, T, CASE, x1, y1, x2, y2, ret, x, y;
char op[5];
int sum[MAXN << 2][MAXN << 2];

void update_r(int cr, int y1, int y2, int rt, int l, int r) {
    if(y1 <= l && r <= y2) {
        sum[cr][rt] = !sum[cr][rt];
        return ;
    }
    int mid = (l + r) >> 1;
    if(y1 <= mid) update_r(cr, y1, y2, lson);
    if(y2 > mid) update_r(cr, y1, y2, rson);
}

void update_c(int x1, int y1, int x2, int y2, int rt, int l, int r) {
    if(x1 <= l && r <= x2) {
        update_r(rt, y1, y2, root);
        return;
    }
    int mid = (l + r) >> 1;
    if(x1 <= mid) update_c(x1, y1, x2, y2, lson);
    if(x2 > mid) update_c(x1, y1, x2, y2, rson);
}
void query_r(int cr, int y1, int y2, int rt, int l, int r) {
    if(sum[cr][rt]) ret ++;
    if(y1 <= l && r <= y2) {
        return ;
    }
    int mid = (l + r) >> 1;
    if(y1 <= mid) query_r(cr, y1, y2, lson);
    if(y2 > mid) query_r(cr, y1, y2, rson);
}

void query_c(int x1, int y1, int x2, int y2, int rt, int l, int r) {
    query_r(rt, y1, y2, root);//选择覆盖了子树的个数
    if(x1 <= l && r <= x2) {
        return;
    }
    int mid = (l + r) >> 1;
    if(x1 <= mid) query_c(x1, y1, x2, y2, lson);
    if(x2 > mid) query_c(x1, y1, x2, y2, rson);
}

int main() {
    //freopen("D://imput.txt","r",stdin);
    scanf("%d", &CASE);
    while(CASE --) {
        memset(sum, 0, sizeof(sum));
        scanf("%d%d", &N, &T);
        while(T --) {
            scanf("%s", op);
            if(op[0] == ‘C‘) {
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                update_c(x1, y1, x2, y2, root);
            } else {
                scanf("%d%d", &x, &y);
                ret = 0;
                query_c(x, y, x, y, root);
                printf("%d\n", ret & 1);
            }
        }
        if(CASE)printf("\n");
    }
    return 0;
}