【poj2155】【Matrix】二位树状数组

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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

题目大意

给一个N*N的矩阵，里面的值不是0，就是1。初始时每一个格子的值为0。

现对该矩阵有两种操作：（共T次）

1.C x1 y1 x2 y2：将左上角为(x1, y1)，右下角为(x2, y2)这个范围的子矩阵里的值全部取反。

2.Q x y：查询矩阵中第i行，第j列的值。

根据数据范围，横纵两个方向都必须是log级的复杂度。如果按照题目原意直接模拟，区间修改单点查询，需要用线段树。但是我并不会二位线段树。那么就利用差分的思想，使其转化为单点修改区间查询，可以用树状数组来维护。

一维的差分是这样的

[le,ri]+val



那么二维的就是



但是详细的，(x,y)+1，是指的从(x,y)到(n,n)的矩阵都+1

那么根据容斥原理



由此一来，单点查询时就查询(0,0)到(x,y)的和

现在就是二维树状数组怎么实现的问题了

其实很简单，就是两个for套在一起就是了

void modify(int x,int y,int val){
for(int i=x;i<=n;i+=(i&(-i)))
for(int j=y;j<=n;j+=(j&(-j)))
c[i][j]++;
}

完整代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=1000+5;
const int T=50000+5;

int n,t;
int c

;

void modify(int x,int y,int val){
for(int i=x;i<=n;i+=(i&(-i)))
for(int j=y;j<=n;j+=(j&(-j)))
c[i][j]++;
}
int query(int x,int y){
int rt=0;
for(int i=x;i>0;i-=(i&(-i)))
for(int j=y;j>0;j-=(j&(-j)))
rt+=c[i][j];
return rt;
}
void solve(){
scanf("%d%d",&n,&t);
memset(c,0,sizeof(c));
while(t--){
char opt[2];
scanf("%s",opt);
if(opt[0]=='C'){
int x1,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
modify(x1,y1,1);
modify(x1,y2+1,-1);
modify(x2+1,y1,-1);
modify(x2+1,y2+1,1);
}
else{
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",query(x,y)%2);
}
}
printf("\n");
}
int main(){
int x;
scanf("%d",&x);
while(x--) solve();
return 0;
}

总结：

1、看到操作不必直接模拟，用差分等思想可以化难为简