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Matrix 二维树状数组的第二类应用

2024-07-13 14:40:31   227人阅读

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 17976 Accepted: 6737

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

#include"iostream"#include"cstdio"#include"cstring"using namespace std;const int ms=1e3+10;int map[ms][ms];int n,m;int lowbit(int t){    return t&(-t);}void updata(int x,int y,int d){    while(x>0)    {        int y1=y;        while(y1>0)        {            map[x][y1]+=d;            y1-=lowbit(y1);        }        x-=lowbit(x);    }}int getsum(int x,int y){    int sum=0;    while(x<=n)    {        int y1=y;        while(y1<=n)        {            sum+=map[x][y1];            y1+=lowbit(y1);        }        x+=lowbit(x);    }    return sum;}int main(){    int i,j,k,t,tcase;    //#ifndef oo    //cout<<"dkkjdkk"<<endl;    //#endif    cin>>tcase;    while(tcase--)    {        cin>>n>>m;        memset(map,0,sizeof(map));        char s[5];//避免回车符的影响  否则  getchar()          while(m--)         {             cin>>s;             if(s[0]==‘C‘)             {                 int x1,y1,x2,y2;                 cin>>x1>>y1>>x2>>y2;                 updata(x2,y2,1);                updata(x1-1,y2,-1);                updata(x1-1,y1-1,1);                updata(x2,y1-1,-1);                    //updata(x1-1,y1-1,1);                 }             else if(s[0]==‘Q‘)             {                 int x,y;                 cin>>x>>y;                 cout<<getsum(x,y)%2<<endl;             }         }         if(tcase)             cout<<endl;    }    return 0;}


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