Matrix
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 30316 | Accepted: 11094 |
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].
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.
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.
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:
给了一个长宽为N的矩阵,C操作可以把以(x1,y1),(x2,y2)为对角线的矩阵内的每一个点进行反转操作。Q是查询(x,y)点的值。
思路:
参考https://wenku.baidu.com/view/1e51750abb68a98271fefaa8。很巧妙的方法。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int sum[1005][1005];
int t,n,m,x1,x2,y1,y2;
int lowbit(int i)
{
return i&(-i);
}
void update(int x,int y){
for(int i=x;i<=n;i+=lowbit(i))
{
for(int j=y;j<=n;j+=lowbit(j))
{
sum[i][j]++;
}
}
}
int query(int x,int y)
{
int ans=0;
for(int i=x;i>0;i-=lowbit(i))
{
for(int j=y;j>0;j-=lowbit(j))
{
ans+=sum[i][j];
}
}
return ans%2;
}
int main()
{
scanf("%d",&t);
while(t--)
{
char s[5];
memset(sum,0,sizeof(sum));
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
scanf("%s",s);
if(s[0]=='C')
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
x1++,x2++,y1++,y2++;
update(x2,y2);
update(x2,y1-1);
update(x1-1,y2);
update(x1-1,y1-1);
}
else
{
scanf("%d%d",&x1,&y1);
printf("%d\n",query(x1,y1));
}
}
printf("\n");
}
return 0;
}