poj 2155 Matrix(二维树状数组，好题)中等难度题目，更新区域，查询单点

1、http://poj.org/problem?id=2155

2、题目大意：

有一个n*n的矩阵，初始值时0，现在对该矩阵做两种操作，C x1 y1 x2 y2，是将这一区域的值是0的换成1,是1的换成0,Q x y查询(x,y)值是多少？

这道题目一看很简单，抬手就写了个for循环，相当于5000*1000*1000的循环，果断超时了，这样的题目就要想到用树状数组来做了，但是这道题目跟以前的树状数组还是不同的，这个是更新一个区域的值，查询一个点的值，想都想不出来哪里可以用树状数组来做，下面看网上大神的思路，很奇妙的。。。

经典的二维树状数组

这道题跟一般的树状数组操作有点不同，它是修改一片区域的值，求的是其中的一个点，我们可以这样来想！！

当我们修改一片区域的时候，我们可以分段去修改，也就是想当于树状数组中的 getsum(x,y)

当我们求其中的一个点的时候，我们可以变为统计这个点的翻转次数，凡是跟这个点相关的区间都要统计一下，

比如说在一维的情况中，

我们要使区间(a, b)内的点 + c，只需要使区间(1, b)内的点+c，而区间(1, a-1)内的点-c即可。

如果是二维的，修改矩阵(x1, y1)到(x2, y2)，即(x2, y2)+c, (x1-1,y2)-c, (x2, y1-1)-c, (x1-1,y1-1)+c。

即我们可以用getsum(x, c)修改(1, x)这个区间内的点的值，而用update(x)来求该点的值

在这里c[]数组记录了翻转次数,只不过题目要求的是01变换,所以c[]只要记录0，1就行了，奇数为1，偶数为0

3、题目：

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 16547 Accepted: 6227

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

POJ Monthly,Lou Tiancheng

 

4、二维树状数组AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 1005int c[N][N];int n;int lowbit(int i){    return i&(-i);}void getsum(int x,int y){    for(int i=x;i>0;i-=lowbit(i))    {        for(int j=y;j>0;j-=lowbit(j))        {            c[i][j]=c[i][j]^1;        }    }}int update(int x,int y){    //sum统计替换的次数，次数是偶数表示输出应该是0,否则是1    int sum=0;    for(int i=x;i<=n;i+=lowbit(i))    {        for(int j=y;j<=n;j+=lowbit(j))        {            sum+=c[i][j];        }    }    if(sum%2==0)    return 0;    else    return 1;}int main(){    int t,m,x1,y1,x2,y2,x,y;    char ch[3];    scanf("%d",&t);    int flag=0;    while(t--)    {        if(flag==0)        {            flag=1;        }        else        printf("\n");        scanf("%d%d",&n,&m);        memset(c,0,sizeof(c));        for(int i=1;i<=m;i++)        {            scanf("%s",ch);            if(ch[0]=='C')            {                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                getsum(x2,y2);                getsum(x1-1,y2);                getsum(x2,y1-1);                getsum(x1-1,y1-1);            }            else            {                scanf("%d%d",&x,&y);                printf("%d\n",update(x,y));            }        }    }    return 0;}/*22 10C 2 1 2 2Q 2 2C 1 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 2 2 2Q 1 1C 1 1 2 1Q 2 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*/

 

5、简单实现超时的代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 1005int a[N][N];int main(){    int t,n,m,x1,y1,x2,y2,x,y;    char ch[3];    scanf("%d",&t);    int flag=0;    while(t--)    {        if(flag!=0)        {            printf("\n");            flag=1;        }        scanf("%d%d",&n,&m);        memset(a,0,sizeof(a));        for(int i=1;i<=m;i++)        {            scanf("%s",ch);            if(ch[0]=='C')            {                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                for(int i=x1;i<=x2;i++)                {                    for(int j=y1;j<=y2;j++)                    {                        a[i][j]=a[i][j]^1;                    }                }            }            else            {                scanf("%d%d",&x,&y);                printf("%d\n",a[x][y]);            }        }    }    return 0;}