poj 2155:Matrix（二维线段树，矩阵取反，好题）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17880		Accepted: 6709

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

POJ Monthly,Lou Tiancheng

---

 

　　二维线段树，矩阵取反，好题。

 

　　题意：

　　

　　思路：

　　矩阵节点的值为是否取反，0为不取反，1为取反，暂称为取反值。

　　取反操作的时候先找到这个矩阵代表的节点，然后将这个节点的值+1再模2，即取反。

　　查询的时候，将（x，y）这个坐标经过的所有矩阵的取反值加起来，每次%2，最后那个值就为这个坐标最后的值。

　　为什么%2，因为不是1就是0，矩阵记录了取反值，找这个坐标的过程中，经过的矩阵如果取反值为1，则结果变为0，在经过一个取反值为1的矩阵，结果又变为1…… 直到加到要找的坐标的取反值，这个结果记录的值就是这个坐标的取反值。这个时候输出结果。

 

　　代码：

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 
 #define MAXN 1100
 
 int tree[MAXN*][MAXN*],s;
 
 void Negate_y(int d,int dy,int L,int R,int y1,int y2)    //取反操作
 {
     if(L==y1 && R==y2){    //将这个矩阵的所有元素记录为取反
         tree[d][dy] = (tree[d][dy]+) % ;
         return ;
     }
 
     int mid = (L+R)>>;
     if(mid>=y2)
         Negate_y(d,dy<<,L,mid,y1,y2);
     else if(mid<y1)
         Negate_y(d,dy<<|,mid+,R,y1,y2);
     else{
         Negate_y(d,dy<<,L,mid,y1,mid);
         Negate_y(d,dy<<|,mid+,R,mid+,y2);
     }
 }
 
 void Negate_x(int d,int L,int R,int x1,int y1,int x2,int y2)        //取反操作
 {
     if(L==x1 && R==x2){    //找到行块
         Negate_y(d,,,s,y1,y2);
         return ;
     }
 
     int mid = (L+R)>>;
     if(mid>=x2)
         Negate_x(d<<,L,mid,x1,y1,x2,y2);
     else if(mid<x1)
         Negate_x(d<<|,mid+,R,x1,y1,x2,y2);
     else{
         Negate_x(d<<,L,mid,x1,y1,mid,y2);
         Negate_x(d<<|,mid+,R,mid+,y1,x2,y2);
     }
 }
 
 int Query_y(int d,int dy,int L,int R,int r)    //查询
 {
     if(L==R)    //找到要找的坐标，输出这个坐标对应的值
         return tree[d][dy];
 
     //没找到
     int mid = (L+R)>>;
     if(mid >= r)
         return (Query_y(d,dy<<,L,mid,r)+tree[d][dy]) % ;
     else
         return (Query_y(d,dy<<|,mid+,R,r)+tree[d][dy]) % ;
 }
 
 int Query_x(int d,int L,int R,int l,int r)    //查询
 {
     if(L==R){    //找到要找的行块，继续查找列块
         return Query_y(d,,,s,r);
     }
 
     //没找到
     int mid = (L+R)>>;
     if(mid >= l)
         return (Query_x(d<<,L,mid,l,r) + Query_y(d,,,s,r)) % ;
     else
         return (Query_x(d<<|,mid+,R,l,r) + Query_y(d,,,s,r)) % ;
 }
 
 int main()
 {
     int X,T,x,y,x1,y1,x2,y2;
     scanf("%d",&X);
     while(X--){
         memset(tree,,sizeof(tree));
         scanf("%d%d",&s,&T);
         while(T--){
             char c[];
             scanf("%s",c);
             if(c[]=='C'){
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 Negate_x(,,s,x1,y1,x2,y2);
             }
             else if(c[]=='Q'){
                 scanf("%d%d",&x,&y);
                 printf("%d\n",Query_x(,,s,x,y));
             }
         }
         if(X!=)
             printf("\n");
     }
 
     return ;
 }

Freecode : www.cnblogs.com/yym2013