模板:
void add(int x,int y,int val)
{
while(y<=n)
{
int tmp=x;
while(tmp<=n)
{
c[tmp][y]+=val;
tmp+=tmp&-tmp;
}
y+=y&-y;
}
}
int getsum(int x,int y)
{
int sum=0;
while(y>0)
{
int tmp=x;
while(tmp>0)
{
sum+=c[tmp][y];
tmp-=tmp&-tmp;
}
y-=y&-y;
}
return sum;
}
对一个矩阵块修改呢要改四个点,下图黄色点是要修改的区域的边界点,我们修改的四个点呢是酱色的外面的大的四个点,原点呢是由题意而定一般是(1,1),(x,y)到(1,1)所有点的和,也就是矩阵块。(图很丑,随便画的凑活着看哈哈),左上和右下是加val,左下和右上是减val.这样求和求出每个点的真正值就不会出错,类似差分数组,存的可以是两个数之间的差值。再累加就是原值,这也是树状数组可以实现区间修改的原理,区间查询也可以,网上很多看一下就懂了。
下面给出一道经典的二维数组应用:
poj2155
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
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].
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
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.
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
There is a blank line between every two continuous test cases.
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 1Sample Output
1 0 0 1
题意:
对矩阵的某一块(矩形)进行取反,很多次取反,最后问现在某个点是0还是1,矩阵仅由0,1构成。
刚开始不知道怎么做,后来看出只有零一的话,不断加一,取反一次就加一,奇数呢证明还是1,偶数就是0,这样就可以用神奇的二维树状数组了,代码如下:
#include<stdio.h>
#include<string.h>
using namespace std;
int c[1002][1002];
int n,t;
void add(int x,int y,int val)
{
while(y<=n)
{
int tmp=x;
while(tmp<=n)
{
c[tmp][y]+=val;
tmp+=tmp&-tmp;
}
y+=y&-y;
}
}
int getsum(int x,int y)
{
int sum=0;
while(y>0)
{
int tmp=x;
while(tmp>0)
{
sum+=c[tmp][y];
tmp-=tmp&-tmp;
}
y-=y&-y;
}
return sum;
}
int main()
{
int x, x1,x2,y1,y2,flag=0;
scanf("%d",&x);
while(x--)
{
memset(c,0,sizeof(c));
if(flag)printf("\n");
flag=1;
scanf("%d%d",&n,&t);
while(t--)
{
char a[3];
scanf("%s",a);
if(a[0]=='C')
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
add(x1,y1,1);
add(x2+1,y1,1);
add(x1,y2+1,1);
add(x2+1,y2+1,1);
}
else
{
scanf("%d%d",&x1,&y1);
printf("%d\n",getsum(x1,y1)%2);
}
}
}
}