poj----2155 Matrix(二维树状数组第二类)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16950		Accepted: 6369

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

 

代码：

采用树状数组第二种方法

采用更新向下，统计向上的方法....楼教主这道题出的还是比较新颖的......

代码：438ms

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 1005
#define lowbit(x) ((x)&(-x))
int aa[maxn][maxn];
int nn;
void ope(int x ,int y ,int val)
 {
   for(int i=x ;i>0 ;i-=lowbit(i))
   {
    for(int j=y ;j>0 ;j-=lowbit(j))
    {
      aa[i][j]+=val;
    }
   }
 }
 int clac(int  x,int y)
 {
   int ans=0;
   for(int i=x;i<=nn ;i+=lowbit(i))
   {
     for(int j=y ;j<=nn ;j+=lowbit(j))
     {
       ans+=aa[i][j];
     }
   }
   return ans;
 }
struct node
 {
   int x;
   int y;
 };

int main()
{
    int tt,xx;
    char str[5];
    node sa,sb;
    scanf("%d",&xx);
    while(xx--)
    {
     memset(aa,0,sizeof(aa));
     scanf("%d%d",&nn,&tt);
     while(tt--)
     {
     scanf("%s",&str);
     if(str[0]=='C')
     {
      scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);
      sa.x--; //左上角全体加1
      sa.y--;
      ope(sb.x,sb.y,1);
      ope(sa.x,sb.y,-1);
      ope(sb.x,sa.y,-1); 
      ope(sa.x,sa.y,1);
     }
    else
    {
     scanf("%d%d",&sa.x,&sa.y);
     printf("%d
",clac(sa.x,sa.y)&1);
    }
   }
   printf("
");
 }
  return 0;
}

改进版..
代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 1005
#define lowbit(x) ((x)&(-x))
int aa[maxn][maxn];
int nn;
void ope(int x ,int y )
 {
   for(int i=x ;i>0 ;i-=lowbit(i))
   {
    for(int j=y ;j>0 ;j-=lowbit(j))
    {
      aa[i][j]=aa[i][j]^1;
    }
   }
 }
 int clac(int  x,int y)
 {
   int ans=0;
   for(int i=x;i<=nn ;i+=lowbit(i))
   {
     for(int j=y ;j<=nn ;j+=lowbit(j))
     {
       ans+=aa[i][j];
     }
   }
   return ans;
 }
struct node
 {
   int x;
   int y;
 };

int main()
{
    int tt,xx;
    char str[5];
    node sa,sb;
    scanf("%d",&xx);
    while(xx--)
    {
     memset(aa,0,sizeof(aa));
     scanf("%d%d",&nn,&tt);
     while(tt--)
     {
     scanf("%s",&str);
     if(str[0]=='C')
     {
      scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);
      sa.x--; //左上角全体加1
      sa.y--;
      ope(sb.x,sb.y);
      ope(sa.x,sb.y);
      ope(sb.x,sa.y); 
      ope(sa.x,sa.y);
     }
    else
    {
     scanf("%d%d",&sa.x,&sa.y);
     printf("%d
",clac(sa.x,sa.y)&1);
    }
   }
   printf("
");
 }
  return 0;
}

 采用树状数组第一种方法

传统的方法：

代码：435ms

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 1005
#define lowbit(x) ((x)&(-x))
int aa[maxn][maxn];
int nn;
void ope(int x ,int y )
 {
   for(int i=x ;i<=nn ;i+=lowbit(i))
     for(int j=y ;j<=nn ;j+=lowbit(j))
         aa[i][j]=aa[i][j]^1;
 }
 int clac(int  x,int y)
 {
   int ans=0,i,j;
   for(i=x;i>0 ;i-=lowbit(i))
     for(j=y ;j>0 ;j-=lowbit(j))
       ans+=aa[i][j];
   return ans;
 }
struct node
 {
   int x,y;
 };
int main()
{
    int tt,xx;
    char str[5];
    node sa,sb;
    scanf("%d",&xx);
    while(xx--)
    {
     memset(aa,0,sizeof(aa));
     scanf("%d%d",&nn,&tt);
     while(tt--)
     {
     scanf("%s",&str);
     if(str[0]=='C')
     {
      scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);
       sb.x++; //左上角全体加1
       sb.y++;
      ope(sb.x,sb.y);
      ope(sa.x,sb.y);
      ope(sb.x,sa.y);
      ope(sa.x,sa.y);
     }
    else
    {
     scanf("%d%d",&sa.x,&sa.y);
     printf("%d
",clac(sa.x,sa.y)&1);
    }
   }
   printf("
");
 }
  return 0;
}