D - Checker
Time limit : 2sec / Memory limit : 256MB
Score : 500 points
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
orW
. - 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
例如,最后一组数据的图为如下(黄色为黑底)
这题做法要比较巧的方法,否则会超时,我的复杂度为O(n+5*k*k),也不知道各位大佬有没有更加快的方法
1
我们可以发现一个完整的黑白相间的图形是由2k为边长的正方形组成的
所以我把所有的点先对他们的x%2k ,y%2k处理一下 ,分不同颜色标记在w[][]和b[][]
这里所花费时间为n
然后dp,w[i][j]代表i*j这个长方形中有多少个白色的点
b[i][j]代表i*j这个长方形中有多少个黑色的点
这里所花费时间为4*k*k
2
我们不知道原点,那就暴力所有可能的情况,你会发现只需要沿x暴力k次,沿y暴力k次,然后我们会发现如下情况(假设k=4)
这是最优状况(黄色不一定代表是黑色的,也有可以是白色的)
这是一般情况
所以可以说s1=五块黑色中所存在的点(可以利用b[][]算出)+四块白色中所存在的点
(反着来) s2=五块白色中所存在的点(可以利用w[][]算出)+四块黑色中所存在的点
不存在的块点数为0
s=Max(s1,s2)
这里所花费时间为k*k
乎,写的时候比较小心,一次AC
#include <cstdio> #include <algorithm> using namespace std; int w[2005][2005],b[2005][2005]; int jisuan(int x1,int y1,int x2,int y2,int t[][2005]){ if(x1>x2||y1>y2) return 0; return t[x2][y2]-t[x1-1][y2]-t[x2][y1-1]+t[x1-1][y1-1]; } int main() { int n,k,x,y; char z; scanf("%d%d",&n,&k); for(int i=1;i<=n;i++){ scanf("%d%d %c",&x,&y,&z); if(z=='W') w[x%(2*k)+1][y%(2*k)+1]++; else b[x%(2*k)+1][y%(2*k)+1]++; } for(int i=1;i<=2*k;i++) for (int j=1;j<=2*k ;j++) { w[i][j]=w[i-1][j]-w[i-1][j-1]+w[i][j-1]+w[i][j]; b[i][j]=b[i-1][j]-b[i-1][j-1]+b[i][j-1]+b[i][j]; } int mmax=0; int t1,t2; for( x=0;x<k;x++) for(y=0;y<k;y++) { t1=jisuan(x+1,y+1,x+k, y+k,w)+jisuan(1,1,x,y,w)+jisuan(x+k+1, y+k+1,2*k,2*k,w)+jisuan(x+k+1, 1,2*k,y,w)+jisuan(1, k+y+1,x,2*k,w);//百色最多五种 t1+=jisuan(1, y+1, x, y+k, b)+jisuan(x+1,1,x+k, y, b)+jisuan(x+k+1, y+1, 2*k, y+k, b)+jisuan(x+1, y+k+1, x+k, 2*k, b);//黑色有4种 //反着来一下 t2=jisuan(x+1,y+1,x+k, y+k,b)+jisuan(1,1,x,y,b)+jisuan(x+k+1, y+k+1,2*k,2*k,b)+jisuan(x+k+1, 1,2*k,y,b)+jisuan(1, k+y+1,x,2*k,b); t2+=jisuan(1, y+1, x, y+k, w)+jisuan(x+1,1,x+k, y, w)+jisuan(x+k+1, y+1, 2*k, y+k, w)+jisuan(x+1, y+k+1, x+k, 2*k, w); mmax=max(t1, max(t2, mmax)); } printf("%d\n",mmax); return 0; }