AtCoder Regular Contest 089 D Checker

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 ≤
4000
 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

Copy

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

Sample Output 1

Copy

4

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

Sample Input 2

Copy

2 1000
0 0 B
0 1 W

Sample Output 2

Copy

2

Sample Input 3

Copy

6 21 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

Copy

4

题目大意：给你一个K*K的无限格子图，你可以自由移动格子图。然后给你N个要求：坐标Xn Yn 处为黑("B")或白("B"),问你有几个请求可以实现。

通过观察可以发现点(x,y)和点(x,y+2k),(x+2k,y)是等价的，要求(x,y)处是某种颜色和在(x,y-k),(x-k,y)处是它的反色等价，读入数据时可以把数据整理下，先整理成一种颜色，然后把x和y全部MOD(2*k)。由于给了两秒，所以第一印象就是穷举所有的(x,y)(0<=x,y<k)点为格子图白色和黑色的右下角起点，判断最多的满足要求数目。有注意到在(x,y)处放黑色块和放白色块也是可以转换的，因为假设放黑色块满足的数目是ans，那么必然有N-ans个请求是不满足的，也就是说这些请求请求的是某色但实际是它的反色。所以当你把放黑块改成放白块后,这些请求就满足了，但之前满足的请求就不满足了。所以对于每一种(x,y)情况取ans和N-ans即可。

然后一波暴力过去，结果超时....

Test case
Case name   Status  Exec time Memory usage
0_000.txt	AC	6 ms	16640 KB
0_001.txt	AC	35 ms	16640 KB
0_002.txt	AC	6 ms	16640 KB
1_003.txt	AC	22 ms	16640 KB
1_004.txt	AC	6 ms	16640 KB
1_005.txt	AC	6 ms	16640 KB
1_006.txt	AC	6 ms	16640 KB
1_007.txt	AC	6 ms	16640 KB
1_008.txt	AC	22 ms	16640 KB
1_009.txt	AC	6 ms	16640 KB
1_010.txt	AC	6 ms	16640 KB
1_011.txt	AC	7 ms	16640 KB
1_012.txt	AC	315 ms	16640 KB
1_013.txt	TLE
1_014.txt	AC	36 ms	17536 KB
1_015.txt	AC	49 ms	17536 KB
1_016.txt	AC	142 ms	17536 KB
1_017.txt	TLE
1_018.txt	TLE
1_019.txt	AC	1459 ms	16768 KB
1_020.txt	AC	1459 ms	16768 KB
1_021.txt	AC	1449 ms	16768 KB
1_022.txt	AC	1453 ms	16768 KB
1_023.txt	AC	1438 ms	16768
9ab1
KB
1_024.txt	AC	1441 ms	16768 KB
1_025.txt	AC	390 ms	17536 KB
1_026.txt	TLE
1_027.txt	AC	358 ms	17536 KB
1_028.txt	TLE
1_029.txt	AC	393 ms	17536 KB
1_030.txt	TLE

超时原因很显然,这种方法是O(N*K*K)的。对于每种情况都要把所有N个请求比较下还是比较蠢的,所以可以一开始就把每个(x,y)到(0,0)的矩形内的要求总数算出来,然后对于每个枚举情况下的(x,y)用类似计算面积的方法统计满足的个数就能直接出答案。

AC代码

#include<bits/stdc++.h>
using namespace std;
int  xx;
int  yy;
int lt[2050][2050];
int main () {
memset(lt,0,sizeof(lt));
int n,k,x,y;
char c[10];
cin>>n>>k;
for(int i=0; i<n; i++) {
scanf("%d %d %s",&xx,&yy,c);
if(c[0]=='W') {
yy+=k;
}
xx%=2*k;
yy%=2*k;
lt[xx][yy]++;
}
for(int i=0; i<=2*k; i++)
for(int j=1; j<=2*k; j++)
lt[i][j]+=lt[i][j-1];
for(int i=1; i<=2*k; i++)
for(int j=0; j<=2*k; j++)
lt[i][j]+=lt[i-1][j];
int  ans=0;
for(int a=1; a<=k; a++) {
for(int b=1; b<=k; b++) {
int c1=lt[b-1][a-1];
int c2=lt[b-1][2*k-1]-lt[b-1][k+a-1];
int c3=lt[2*k-1][a-1]-lt[k+b-1][a-1];
int c4=lt[2*k-1][2*k-1]-lt[k+b-1][2*k-1]-lt[2*k-1][k+a-1]+lt[k+b-1][a+k-1];
int c5=lt[k+b-1][a+k-1]-lt[b-1][a+k-1]-lt[b+k-1][a-1]+lt[b-1][a-1];
ans=max(ans,c1+c2+c3+c4+c5);
ans=max(ans,n-(c1+c2+c3+c4+c5));
}
}
printf("%d\n",ans);
return 0;
}

Test case
Case name   Status Exec time Memory usage
0_000.txt	AC	5 ms	16640 KB
0_001.txt	AC	29 ms	16640 KB
0_002.txt	AC	5 ms	16640 KB
1_003.txt	AC	29 ms	16640 KB
1_004.txt	AC	5 ms	16640 KB
1_005.txt	AC	5 ms	16640 KB
1_006.txt	AC	5 ms	16640 KB
1_007.txt	AC	5 ms	16640 KB
1_008.txt	AC	30 ms	16640 KB
1_009.txt	AC	6 ms	16640 KB
1_010.txt	AC	6 ms	16640 KB
1_011.txt	AC	6 ms	16640 KB
1_012.txt	AC	6 ms	16640 KB
1_013.txt	AC	30 ms	16640 KB
1_014.txt	AC	32 ms	16640 KB
1_015.txt	AC	32 ms	16640 KB
1_016.txt	AC	32 ms	16640 KB
1_017.txt	AC	33 ms	16640 KB
1_018.txt	AC	57 ms	16640 KB
1_019.txt	AC	8 ms	16640 KB
1_020.txt	AC	7 ms	16640 KB
1_021.txt	AC	8 ms	16640 KB
1_022.txt	AC	7 ms	16640 KB
1_023.txt	AC	8 ms	16640 KB
1_024.txt	AC	8 ms	16640 KB
1_025.txt	AC	28 ms	16640 KB
1_026.txt	AC	53 ms	16640 KB
1_027.txt	AC	28 ms	16640 KB
1_028.txt	AC	52 ms	16640 KB
1_029.txt	AC	28 ms	16640 KB
1_030.txt	AC	52 ms	16640 KB