http://poj.org/problem?id=2155
Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 18769 | Accepted: 7078 |
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
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);
画个图想想就知道了。
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;
}