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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]={};//全部的數字有幾種
  7. int sub[301]={};
  8. int nowN;//現在的數字有幾種
  9.  
  10. void recount(int lx,int ly){
  11. 	for(int n=0; n<=N; n+=1)
  12. 		sub[n]=0;
  13. 	for(int dx=0; dx<r; dx+=1)
  14. 		for(int dy=0; dy<c; dy+=1){
  15. 			int color=num[ lx+dx ][ ly+dy ];
  16. 			sub[ color ]+=1;
  17. 			nowN-= sub[ color ]==all[ color ];
  18. 		}
  19. }
  20. int main(){
  21. 	int allN=0;
  22.  
  23. 	cin>>R>>C>>N>>r>>c;
  24. 	for(int x=0; x<R; x+=1)
  25. 		for(int y=0; y<C; y+=1){
  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. 		// sliding window
  37. 		for(int Ly=1; Ly<=C-c; Ly+=1){
  38. 			// 讓最左邊的 column 露出來
  39. 			for(int dx=0; dx<r; dx+=1){
  40. 				int color=num[Lx+dx][Ly-1];
  41. 				nowN+= sub[color]==all[color];
  42. 				sub[color]-=1;
  43. 			}
  44. 			// 讓最右邊的 column 遮起來
  45. 			for(int dx=0; dx<r; dx+=1){
  46. 				int color=num[Lx+dx][Ly+c-1];
  47. 				sub[color]+=1;
  48. 				nowN-= sub[color]==all[color];
  49. 			}
  50. 			cout<<" "<<nowN;
  51. 		}
  52. 		cout<<"\n";
  53. 	}
  54.  
  55. }
Success #stdin #stdout 0.01s 5280KB
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https://ideone.com/sS4T2R
language:
C++ (gcc 8.3)
created:
33 days ago
visibility:
public

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