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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. int R,C,N,r,c;
  5. int num[300][300];
  6. int all[301]={};
  7. int sub[301]={};
  8. int nowN;
  9. void recount(int lx, int ly){
  10. 	for(int n=0;n<=N;n++)
  11. 		sub[n]=0;
  12. 	for(int dx=0;dx<r;dx++){
  13. 		for(int dy=0;dy<c;dy++){
  14. 			int color=num[lx+dx][ly+dy];
  15. 			sub[color]+=1;
  16. 			nowN-=sub[color]==all[color];
  17. 		}
  18. 	}
  19. }
  20.  
  21. int main() {
  22. 	int allN=0;
  23. 	cin>>R>>C>>N>>r>>c;
  24. 	for(int x=0;x<R;x++){
  25. 		for(int y=0;y<C;y++){
  26. 			cin>>num[x][y];
  27. 			allN+=all[num[x][y]]==0;
  28. 			all[num[x][y]]+=1;
  29. 		}
  30. 	}
  31. 	for(int Lx=0;Lx<=R-r;Lx+=1){
  32. 		//mask 對齊最左側時才需要重算
  33. 		nowN=allN;
  34. 		recount(Lx,0);
  35. 		cout<<nowN;
  36. 		for(int Ly=1;Ly<=C-c;Ly++){
  37. 			//讓最左邊的一列露出來、最右邊的遮起來
  38. 			for(int dx=0;dx<r;dx++){
  39. 				int color =num[Lx+dx][Ly-1];
  40. 				nowN+=sub[color]==all[color];
  41. 				sub[color]-=1;
  42. 			}
  43. 			for(int dx=0; dx<r; dx+=1){
  44. 				int color=num[Lx+dx][Ly+c-1];
  45. 				sub[color]+=1;
  46. 				nowN-= sub[color]==all[color];
  47. 			}
  48. 			cout<<" "<<nowN;
  49. 		}
  50. 		cout<<"\n";
  51. 	}
  52. 	return 0;
  53. }
Success #stdin #stdout 0.01s 5304KB
comments (?)
stdin
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5 6 9 3 4
7 1 5 3 9 5
4 5 4 5 1 2
6 1 6 2 9 7
4 7 1 5 8 8
3 4 3 3 5 3
stdout
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8 8 7
8 9 7
8 9 8
https://ideone.com/aKRKXH
278e
language:
C++14 (gcc 8.3)
created:
33 days ago
visibility:
public

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