poj 2155 Matrix 二维树状数组

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Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 19691 Accepted: 7387

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

题意：给一个N*N的矩阵，里面的值不是0，就是1。初始时每一个格子的值为0。
现对该矩阵有两种操作：（共T次）
1.C x1 y1 x2 y2：将左上角为(x1, y1)，右下角为(x2, y2)这个范围的子矩阵里的值全部取反。
2.Q x y：查询矩阵中第i行，第j列的值。

国家队集训队论文：武森《浅谈信息学竞赛中的“0”和“1”》

1. 根据这个题目中介绍的这个矩阵中的数的特点不是 1 就是 0，这样我们只需记录每个格子改变过几次，即可判断这个格子的数字。
2. 先考虑一维的情况：
若要修改[x,y]区间的值，其实可以先只修改 x 和 y+1 这两个点的值（将这两个点的值加1）。查询k点的值时，其修改次数即为 sum(cnt[1] + … + cnt[k])。
3. 二维的情况：
道理同一维。要修改范围[x1, y1, x2, y2]，只需修改这四个点：(x1,y1), (x1,y2+1), (x2+1,y1), (x2+1,y2+1)。查询点(x,y)的值时，其修改次数为 sum(cnt[1, 1, x, y])。
4. 而区间求和，便可用树状数组来实现。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#define MAXN 1111using namespace std;int bit[MAXN][MAXN];int lowbit(int x){    return x&(-x);}void add(int x,int y,int num){    for(int i=x;i<MAXN;i+=lowbit(i))        for(int j=y;j<MAXN;j+=lowbit(j))            bit[i][j]+=num;}int getsum(int x,int y){    int cnt=0;    for(int i=x;i>0;i-=lowbit(i))        for(int j=y;j>0;j-=lowbit(j))            cnt+=bit[i][j];    return cnt;}int main(){    int t,n,m;    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&m);        memset(bit,0,sizeof(bit));        while(m--){            char op[2];            scanf("%s",op);            if(op[0]=='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);            }            else {                int x,y;                scanf("%d %d",&x,&y);                printf("%d\n",getsum(x,y)%2);            }        }        puts("");    }    return 0;}