poj2155 Matrix 二维数组数组

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]. 

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. 

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. 

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

1
0
0
1

POJ Monthly,Lou Tiancheng

题意：二维矩阵，可以把子矩阵反转，查询某点的状态。

题解：二维树状数组可搞。

属于树状数组的第二类应用，区间修改单点查值。

（2）“改段求点”型，即对于序列A有以下操作：

【1】修改操作：将A[l..r]之间的全部元素值加上c；

【2】求和操作：求此时A[x]的值。

这个模型中需要设置一个辅助数组B：B[i]表示A[1..i]到目前为止共被整体加了多少（或者可以说成，到目前为止的所有ADD(i, c)操作中c的总和）。

则可以发现，对于之前的所有ADD(x, c)操作，当且仅当x>=i时，该操作会对A[i]的值造成影响（将A[i]加上c），又由于初始A[i]=0，所以有A[i] = B[i..N]之和！而ADD(i, c)（将A[1..i]整体加上c），将B[i]加上c即可——只要对B数组进行操作就行了。

这样就把该模型转化成了“改点求段”型，只是有一点不同的是，SUM(x)不是求B[1..x]的和而是求B[x..N]的和，此时只需把ADD和SUM中的增减次序对调即可（模型1中是ADD加SUM减，这里是ADD减SUM加）。代码：

void ADD(int x, int c)
{
     for (int i=x; i>0; i-=i&(-i)) b[i] += c;
}
int SUM(int x)
{
    int s = 0;
    for (int i=x; i<=n; i+=i&(-i)) s += b[i];
    return s;
}

操作【1】：ADD(l-1, -c); ADD(r, c);

操作【2】：SUM(x)。

对于这道题，就是变为二维的而已，取反操作可以看成+1,最后就看是否整除2即可。

比如操作x1,y1,x2,y2的子矩形，就是

 add(x2,y2,1);
 add(x1-1,y2,-1);
 add(x2,y1-1,-1);
 add(x1-1,y1-1,1);

代码：

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=1005;
int bit[N][N];
int n;
int sum(int x,int y)
{
    int s=0;
    for(int i=x;i<=n;i+=i&-i)
        for(int j=y;j<=n;j+=j&-j)
            s+=bit[i][j];
    return s;
}
void add(int x,int y,int val)
{
    for(int i=x;i>0;i-=i&-i)
        for(int j=y;j>0;j-=j&-j)
            bit[i][j]+=val;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int q;
        scanf("%d%d",&n,&q);
        clr(bit);
        while(q--)
        {
            char s[5];
            scanf("%s",s);
            int x1,x2,y1,y2;
            if(s[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x2,y2,1);
                add(x1-1,y2,-1);
                add(x2,y1-1,-1);
                add(x1-1,y1-1,1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1)%2);
            }
        }
        puts("");
    }
    return 0;
}