Matrix
| Time Limit: 3000MS |
|
Memory Limit: 65536K |
| Total Submissions: 17766 |
|
Accepted: 6674 |
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
题意：对一个给定size且初始化为0的矩阵，执行一些命令，Q A B为查看arr[a][b]元素的值，C X1 Y1 X2 Y2为将(x1, y1) (x2, y2)矩形范围内的所有点0、1翻转。
题解：树状数组模式二的用法，段更新，点查询。update(x2, y2)表示从(1, 1)到(x2, y2)范围内的所有点都要翻转一次，但是这样会把给定范围外的一些点也翻转到，因此需要将这些点翻转回去。
#include <stdio.h>
#include <string.h>
#define maxn 1002
int size, tree[maxn][maxn];
int lowBit(int x){ return x & (-x); }
//向下更新表示A[1]...A[i]每个元素都要 += val,推广到二维同理
void update(int x, int y, int val)
{
int temp;
while(x > 0){
temp = y;
while(temp > 0){
tree[x][temp] += val;
temp -= lowBit(temp);
}
x -= lowBit(x);
}
}
int query(int x, int y)
{
int sum = 0, temp;
while(x <= size){
temp = y;
while(temp <= size){
sum += tree[x][temp];
temp += lowBit(temp);
}
x += lowBit(x);
}
return sum;
}
int main()
{
//freopen("stdin.txt", "r", stdin);
int cas, q, a, b, c, d;
char com[2];
scanf("%d", &cas);
while(cas--){
scanf("%d%d", &size, &q);
memset(tree, 0, sizeof(tree));
while(q--){
scanf("%s%d%d", com, &a, &b);
if(com[0] == 'C'){
scanf("%d%d", &c, &d);
update(c, b - 1, -1);
update(a - 1, d, -1);
update(a - 1, b - 1, 1);
update(c, d, 1);
}else printf("%d\n", query(a, b) & 1);
}
if(cas) printf("\n");
}
return 0;
}
POJ2155 Matrix 【二维树状数组】+【段更新点查询】,布布扣,bubuko.com
POJ2155 Matrix 【二维树状数组】+【段更新点查询】
原文：http://blog.csdn.net/chang_mu/article/details/37773141
