Matrix
Time Limit: 3000 MS Memory Limit: 65536 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define max 1005
int c[max][max];
int n;
int lowbit(int x)
{
return x&-x;
}
void update(int x,int y,int val)
{
for(int i=x; i<=n; i+=lowbit(i))
{
for(int j=y; j<=n; j+=lowbit(j))
{
c[i][j]+=val;
}
}
}
int get_sum(int x,int y)
{
int s=0;
for(int i=x; i>0; i-=lowbit(i))
{
for(int j=y; j>0; j-=lowbit(j))
{
s+=c[i][j];
}
}
return s;
}
int main()
{
/// freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);
int cc,t,x1,x2,y1,y2,a,b;
char ch;
scanf("%d",&cc);
for(int i=1;i<=cc;i++)
{
memset(c,0,sizeof(c));
scanf("%d%d",&n,&t);
getchar();
for(int j=0;j<t;j++) ///下标不可以从0开始 会死循环
{
scanf("%c",&ch);
if(ch==‘C‘)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
getchar();
update(x1,y1,1);
update(x1,y2+1,1);
update(x2+1,y1,1);
update(x2+1,y2+1,1);
}
else
{
scanf("%d%d",&a,&b);
getchar();
printf("%d\n",get_sum(a,b)%2);
}
}
printf("\n");
}
return 0;
}
* 矩阵的00在左上角
http://blog.csdn.net/zxy_snow/article/details/6264135 图示
poj 2155 区间更新 单点查询