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*3][MAXN*3],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]+1) % 2;
        return ;
    }

    int mid = (L+R)>>1;
    if(mid>=y2)
        Negate_y(d,dy<<1,L,mid,y1,y2);
    else if(mid<y1)
        Negate_y(d,dy<<1|1,mid+1,R,y1,y2);
    else{
        Negate_y(d,dy<<1,L,mid,y1,mid);
        Negate_y(d,dy<<1|1,mid+1,R,mid+1,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,1,1,s,y1,y2);    
        return ;
    }

    int mid = (L+R)>>1;
    if(mid>=x2)
        Negate_x(d<<1,L,mid,x1,y1,x2,y2);
    else if(mid<x1)
        Negate_x(d<<1|1,mid+1,R,x1,y1,x2,y2);
    else{
        Negate_x(d<<1,L,mid,x1,y1,mid,y2);
        Negate_x(d<<1|1,mid+1,R,mid+1,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)>>1;
    if(mid >= r)
        return (Query_y(d,dy<<1,L,mid,r)+tree[d][dy]) % 2;
    else
        return (Query_y(d,dy<<1|1,mid+1,R,r)+tree[d][dy]) % 2;
}

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,1,1,s,r);
    }

    //没找到
    int mid = (L+R)>>1;
    if(mid >= l)
        return (Query_x(d<<1,L,mid,l,r) + Query_y(d,1,1,s,r)) % 2;
    else
        return (Query_x(d<<1|1,mid+1,R,l,r) + Query_y(d,1,1,s,r)) % 2;
}

int main()
{
    int X,T,x,y,x1,y1,x2,y2;
    scanf("%d",&X);
    while(X--){
        memset(tree,0,sizeof(tree));
        scanf("%d%d",&s,&T);
        while(T--){
            char c[5];
            scanf("%s",c);
            if(c[0]=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                Negate_x(1,1,s,x1,y1,x2,y2);
            }
            else if(c[0]=='Q'){
                scanf("%d%d",&x,&y);
                printf("%d
",Query_x(1,1,s,x,y));
            }
        }
        if(X!=0)
            printf("
");
    }

    return 0;
}

Freecode : www.cnblogs.com/yym2013