（二维线段树） Matrix (P2155)

这个题可以更简单快捷地用二维数状数组解出。，自己写了很多，各种冲突，还是先转一个吧，

#include   
    #include   
    #include   
    using namespace std;  
    const int N=1010;  
    int n,m,ans;  
    struct newtree{  
        int left,right,value;  
        int getnewmid(){  
           return (left+right)/2;  
        }  
    };  
    struct tree{  
        int left,right;  
        newtree newtt[4*N];  
        int getmid(){  
          return (left+right)/2;  
        }  
    }tt[4*N];  
    void built_treey(int lp,int rp,int posx,int posy){  
        tt[posx].newtt[posy].left=lp;  
        tt[posx].newtt[posy].right=rp;  
        tt[posx].newtt[posy].value=0;  
        if(lp==rp)return;  
        int mid=tt[posx].newtt[posy].getnewmid();  
        built_treey(lp,mid,posx,posy*2);  
        built_treey(mid+1,rp,posx,posy*2+1);  
    }  
    void built_treex(int lp,int rp,int pos){  
        tt[pos].left=lp;  
        tt[pos].right=rp;  
        built_treey(1,n,pos,1);  
        if(lp==rp)return;  
        int mid=tt[pos].getmid();  
        built_treex(lp,mid,pos*2);  
        built_treex(mid+1,rp,pos*2+1);  
    }  
    void update_y(int y1,int y2,int posx,int posy){  
        if(tt[posx].newtt[posy].left==y1&&tt[posx].newtt[posy].right==y2){  
          tt[posx].newtt[posy].value=!tt[posx].newtt[posy].value;  
          return;  
        }  
        int mid=tt[posx].newtt[posy].getnewmid();  
        if(y2<=mid)  
            update_y(y1,y2,posx,posy*2);  
        else if(y1>mid)  
            update_y(y1,y2,posx,posy*2+1);  
        else{  
          update_y(y1,mid,posx,posy*2);  
          update_y(mid+1,y2,posx,posy*2+1);  
        }  
    }  
    void update_x(int x1,int x2,int y1,int y2,int pos){  
        if(tt[pos].left==x1&&tt[pos].right==x2){  
          update_y(y1,y2,pos,1);  
          return;  
        }  
        int mid=tt[pos].getmid();  
        if(x2<=mid)  
            update_x(x1,x2,y1,y2,pos*2);  
        else if(x1>mid)  
            update_x(x1,x2,y1,y2,pos*2+1);  
        else{  
           update_x(x1,mid,y1,y2,pos*2);  
           update_x(mid+1,x2,y1,y2,pos*2+1);  
        }  
    }  
    void find_y(int y,int posx,int posy){  
        ans^=tt[posx].newtt[posy].value;  
        if(tt[posx].newtt[posy].left==tt[posx].newtt[posy].right)  
            return;  
        int mid=tt[posx].newtt[posy].getnewmid();  
        if(y>mid)  
            find_y(y,posx,posy*2+1);  
        else  
            find_y(y,posx,posy*2);  
    }  
    void find_x(int x,int y,int pos){  
        find_y(y,pos,1);  
        if(tt[pos].left==tt[pos].right)  
            return;  
        int mid=tt[pos].getmid();  
        if(x>mid)  
            find_x(x,y,pos*2+1);  
        else  
            find_x(x,y,pos*2);  
    }  
    int main(){  
        //freopen("1.txt","r",stdin);  
        int numcase;  
        scanf("%d",&numcase);  
        while(numcase--){  
          scanf("%d%d",&n,&m);  
          built_treex(1,n,1);  
          char ss[2];  
          int x1,x2,y1,y2,x,y;  
          while(m--){  
            scanf("%s",ss);  
            if(ss[0]=='C'){  
              scanf("%d%d%d%d",&x1,&y1,&x2,&y2);  
              update_x(x1,x2,y1,y2,1);  
            }  
            else{  
              scanf("%d%d",&x,&y);  
              ans=0;  
              find_x(x,y,1);  
              printf("%d\n",ans);  
            }  
          }  
          puts("");  
        }  
        return 0;  
    }  

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12243		Accepted: 4627

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

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

Source