Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25843		Accepted: 9545

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

,Lou Tiancheng

 

二维线段树

区间修改，单点查询。

修改时，区间内0改成1,1改成0。这里等价成每次值+1，求解时%2

二维线段树的标记无法下传，解决方法是标记永久化，求值时，累积树上每一层的标记。

RE了好久，发现是边界写错了……

1 #include
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 #define ls l,mid,rt<<1 6 #define rs mid+1,r,rt<<1|1 7 using namespace std; 8 const int mxn=4020; 9 int t[mxn][mxn];10 int n,m;11 int read(){12     int x=0,f=1;char ch=getchar();13     while(ch<'0' || ch>'9'){
   if(ch=='-')f=-1;ch=getchar();}14     while(ch>='0' && ch<='9'){x=x*10-'0'+ch;ch=getchar();}15     return x*f;16 }17 void updateY(int L,int R,int v,int l,int r,int rt,int px){18 //    printf("UY:L:%d R:%d  v:%d   %d %d %d %d\n",L,R,v,l,r,rt,px);19     if(L<=l && r<=R){20         t[px][rt]+=v;return;21     }22     int mid=(l+r)>>1;23     if(L<=mid)updateY(L,R,v,ls,px);24     if(R>mid)updateY(L,R,v,rs,px);25     return;26 }27 void updateX(int L,int R,int yy1,int yy2,int v,int l,int r,int rt){28     if(L<=l && r<=R){29         updateY(yy1,yy2,v,1,n,1,rt);return;30     }31     int mid=(l+r)>>1;32     if(L<=mid)updateX(L,R,yy1,yy2,v,ls);33     if(R>mid)updateX(L,R,yy1,yy2,v,rs);34      return;35 }36 int qY(int y,int l,int r,int rt,int px){37     if(l==r){
   return t[px][rt];}38     int mid=(l+r)>>1;39     if(y<=mid)return t[px][rt]+qY(y,ls,px);40     else return t[px][rt]+qY(y,rs,px);41 }42 int qX(int x,int y,int l,int r,int rt){43     int res=qY(y,1,n,1,rt);44     if(l==r)return res;45     int mid=(l+r)>>1;46     if(x<=mid)return res+qX(x,y,ls);47     else return res+qX(x,y,rs);48 }49 int main(){50     int T=read();51     int i,j;52     int x1,x2,yy1,yy2;53     while(T--){54         memset(t,0,sizeof t);55         //56         n=read();m=read();57         char op[3];58         while(m--){59             scanf("%s",op);60             if(op[0]=='C'){61                 x1=read(),yy1=read(),x2=read(),yy2=read();62                 updateX(x1,x2,yy1,yy2,1,1,n,1);63             }64             else{65                 int x1=read(),yy1=read();66                 int ans=qX(x1,yy1,1,n,1);67                 if(ans&1)printf("1\n");68                 else printf("0\n");69             }70         }71         printf("\n");72     }73     return 0;74 }