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poj 2155 二维树状数组

2024-07-11 17:10:23   226人阅读

http://poj.org/problem?id=2155

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 17721 Accepted: 6653

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

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

100
１


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二维树状数组题，挺难理解的，楼教主的经典题啊，还是要慢慢才能理解
参考下这个链接：http://blog.csdn.net/zxy_snow/article/details/6264135

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#define Maxx 1005int n,m;int c[Maxx][Maxx];int Lowbit(int x){    return x&(-x);}void Update(int x,int y,int i){    int y1;    while(x<=n)    {        y1=y;        while(y1<=n)        {            c[x][y1]+=i;            y1+=Lowbit(y1);           }        x+=Lowbit(x);     }}int GetSum(int x,int y){    int s=0;    while(x>0)        {        int y1=y;            while(y1>0)                {                    s+=c[x][y1];                    y1-=Lowbit(y1);                }            x-=Lowbit(x);        }    return s;}int main(){    int i,j,t;    char cas;    int x1,y1,x,y,a,b;    scanf("%d",&t);    while(t--)        {            memset(c,0,sizeof(c));            scanf("%d%d%*c",&n,&m);            while(m--)                {                    scanf("%c",&cas);                    if(cas == ‘C‘)                        {                            scanf("%d%d%d%d%*c",&x,&y,&x1,&y1);                            x++;y++;x1++;y1++;                            Update(x1,y1,1);                            Update(x-1,y1,1);                            Update(x1,y-1,1);                            Update(x-1,y-1,1);                        }                    else                        {                            scanf("%d%d%*c",&a,&b);                            printf("%d\n",GetSum(a,b)%2);                        }                }            printf("\n");        }    return 0;}

 


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