POJ 2155 Matrix

二维树状数组，可以快速求一个矩阵内的数字之和，类推三位树状数组，快速求一个长方体内的数字和？

POJ - 2155

Matrix

Time Limit: 3000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

Submit Status

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

1001

Source

POJ Monthly,Lou Tiancheng

Submit Status

Problem descriptions:

System Crawler 2016-05-28 0

Initialization.

#include <stdio.h>#include <iostream>#include<algorithm>#include <string.h>using namespace std;typedef long long LL;#define N 1000int tree[N+5][N+5],n,t;int lowbit(int x) {    return x&-x;}void add(int x,int y) {    int t=y;    while(x<=n) {        y=t;        while(y<=n) {            tree[y][x]^=1;            y+=lowbit(y);        }        x+=lowbit(x);    }}int Sum(int x,int y) {    int tot=0,t=y;    while(x) {            y=t;        while(y) {            tot^=tree[y][x];            y-=lowbit(y);        }        x-=lowbit(x);    }    return tot;}int main() {    //freopen("input.txt","r",stdin);    int T,a,b,c,d;    char f;    scanf("%d",&T);    while(T--) {        scanf("%d%d",&n,&t);        for(int i=0; i<t; i++) {            getchar();            scanf("%c",&f);            if(f=='C') {                scanf("%d%d%d%d",&a,&b,&c,&d);                c++;                d++;                add(a,b);                add(c,b);                add(c,d);                add(a,d);            } else {                scanf("%d%d",&a,&b);                printf("%d\n",Sum(a,b));            }        }        printf("\n");        memset(tree,0,sizeof(tree));    }    return 0;}