Checker【解题报告】

Checker —AtCoder - 3876

Problem Statement

AtCoDeer is thinking of painting an infinite two-dimensional grid in a checked pattern of side K. Here, a checked pattern of side K is a pattern where each square is painted black or white so that each connected component of each color is a K × K square. Below is an example of a checked pattern of side 3:

AtCoDeer has N desires. The i-th desire is represented by xi, yi and ci. If ci is B, it means that he wants to paint the square (xi,yi) black; if ci is W, he wants to paint the square (xi,yi) white. At most how many desires can he satisfy at the same time?

Constraints

1 ≤ N ≤ 105
1 ≤ K ≤ 1000
0 ≤ xi ≤ 109
0 ≤ yi ≤ 109
If i ≠ j, then (xi,yi) ≠ (xj,yj).
ci is B or W.
N, K, xi and yi are integers.

Input

Input is given from Standard Input in the following format:

N K
x1 y1 c1
x2 y2 c2
:
xN yN cN

Output

Print the maximum number of desires that can be satisfied at the same time.

Sample Input 1

4 3
0 1 W
1 2 W
5 3 B
5 4 B

Sample Output 1

4

He can satisfy all his desires by painting as shown in the example above.

Sample Input 2

2 1000
0 0 B
0 1 W

Sample Output 2

2

Sample Input 3

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

4

解析

啦啦啦，面对这道题我们首先要明白题意（就是因为题意理解错误，害我废了多少时间……）：此题大概讲述的是，有一张无限大的’棋盘’，然后那张棋盘里包括了无数的k*k的小的方框，求一个原点，使得与规定的点的坐标尽量相同。
这道题，我们可以看出：它并不是一道算法题，那么，我们应该怎么思考呢？大体思路如下：
①我们可以看到，这道题的规模还是蛮大的，所以我们首先要做的是降维和移位，当然还有将字符转化为数字组成；
移位：

不难想到，其实移位就是平移：将每一个点的坐标%（2k），但是这不会影响结果吗？答案是不会的，因为*2k或%2k都仅仅是改变了它的位置，而没有改变它的颜色。

如此，我们便可以进行平移的行动，其次，是降维，这个将会在代码中体现。而将字符转化为数字，则单独标记处理即可。
②我们可以通过移动的方式来实现、找到到底有多少个点使我们所需要的。这可以分为两个部分：一是通过移动移动’x轴’（其实是平行于x轴）或’y轴’（其实是平行于y轴）上的线段来不断的’枚举‘（如上图的黄色部分）。二是通过平移’x轴’向上，一则减，一则加。
【代码实现】

#include<cstdio>
#include<cstring>
const int MAXN = 100000;
const int MAXK = 1000;
int x[MAXN + 5], y[MAXN + 5], c[MAXN + 5];
int y1[10 * MAXK + 5];
int main() {
    int N, K, Max = 0, cnt = 0;
    scanf("%d%d",&N,&K);
    for(int i=1;i<=N;i++) {
        char ch;
        scanf("%d%d %c",&x[i],&y[i],&ch);
        x[i] = x[i] % (2*K);//降维 
        y[i] = y[i] % (2*K);
        if( ch == 'W' ) c[i] = 1;
        else c[i] = 0;
    }
    for(int i=0;i<=K;i++) 
    {
        cnt = 0;
        memset(y1,0,sizeof(y1));
        for(int j=1;j<=N;j++) 
        {
            int p = ( i<=x[j] && x[j] < i+K );
            int q = ( 0 <=y[j] && y[j] < K );
            //printf("%d %d",p,q); 
            if( p ^ q == c[j] )//通过位与来实现
            {
                cnt++;
                y1[y[j]] ++;
            }
            else 
            {
                y1[y[j]] --;
            }
        }
        for(int j=0;j<=K;j++) 
        {
            if( cnt > Max ) Max = cnt;
            if( N - cnt > Max ) Max = N - cnt;
            if( j != K )
                cnt = cnt - y1[j] - y1[K+j];
        }
    }
    printf("%d\n",Max);
}