Poj2155 Matrix 树状数组
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题目描述:
Matrix
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 26468 Accepted: 9778
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].
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.
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.
There is a blank line between every two continuous test cases.
Sample Input
12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1
Sample Output
1001
Source
POJ Monthly,Lou Tiancheng
思路:
因为是改变一个矩阵求点,可以联想到树状数组的该段求点的算法。
只用加一层循环,相当于横着竖着都是一个一维树状数组。
关于add函数:
改变矩阵(x1,x2,y1,y2)所需的操作就是add(x1,y1,1),add(x1,y2+1,-1),add(x2+1,y1,-1),add(x2+1,y2+1,1)。
代码:
#include<cstdio>#include<iostream>#include<cstring>using namespace std;int n,m;int a[1010][1010]= {0};int lowbit(int x) {return x&(-x);}void add(int x,int y,int z) {for(int i=x; i<=n; i+=lowbit(i))for(int j=y; j<=n; j+=lowbit(j))a[i][j]+=z;}int getsum(int x,int y) {int s=0;for(int i=x; i>0; i-=lowbit(i)) {for(int j=y; j>0; j-=lowbit(j)) {s+=a[i][j];}}return s;} int main() {int t;scanf("%d",&t);while(t--) {memset(a,0,sizeof(a));scanf("%d%d",&n,&m);for(int i=1; i<=m; i++) {char ask;scanf("%*c%c",&ask);if(ask=='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);}if(ask=='Q') {int x,y;scanf("%d%d",&x,&y);printf("%d\n",getsum(x,y)%2);}}printf("\n");}return 0;}
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