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

Sample Output

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