Test instructions

Give a diagram of how many squares of different sizes or positions are different.

Analysis:

This kind of question has the idea, the first idea is enumerates the square position, needs the double circulation, the enumeration side long one heavy circulation, the judgment is the square and needs a heavy circulation, the complexity is O (N4), for n≤9, this complexity is acceptable.

You can use the time of O (1) to determine whether it is a square, as in the preprocessing prefix and so the total complexity is optimized to O (N3).

This method is transferred from here

We can think that vertical or horizontal lines is edges between-adjecent point. After, we can take a three dimensional array (say a [n][n][2]) to store the count of horizontal (a[i][j][0]) edges and Vertical (a[i][j][1]) edges. A[i][j][0] contains number of horizontal edges at row i upto Coloumn J. and A[i][j][1] contains number of vertical edges a T Coloumn j upto Row I. Next You use a O (n^2) loop to find a square. A square of size 1 is found if there are an edge from (i,j) to (i,j+1) and (i,j+1) to (i+1,j+1) and (I,j) to (I+1,J) and (i +1,J) to (i+1,j+1) we can get the just by subtracting values calculated above.

For example, A[i][j][0] indicates that on line I, from the first column to the number of horizontal sides of column J, if A[I][J+L][0]-a[i][j][0], the description point (i, J) to (I, j+l) has a horizontal line length of L.

I was also entered into the pit, notice which number represents the row and which number represents the column after VH.

1#include <cstdio>2#include <cstring>3 4 Const intMAXN =Ten;5 BOOLg[2][MAXN][MAXN];6 inta[2][MAXN][MAXN], CNT[MAXN];7 8 intMain ()9 {Ten //freopen ("In.txt", "R", stdin); One intN, m, Kase =0; A while(SCANF ("%d", &n) = =1&&N) - { -memset (G,false,sizeof(G)); theMemset (A,0,sizeof(a)); -memset (CNT,0,sizeof(CNT)); -scanf"%d", &m); - GetChar (); + for(intK =0; K < M; ++k) - { + CharC; A intI, J; atscanf"%c%d%d", &c, &i, &j); - GetChar (); - if(c = ='H') g[0][i][j+1] =true; - Elseg[1][j+1][i] =true; - } - for(inti =1; I <= N; ++i) in for(intj =1; J <= N; ++j) - { toa[0][I][J] = a[0][i][j-1] + g[0][i][j]; +a[1][I][J] = a[1][i-1][J] + g[1][i][j]; - } the * for(inti =1; I < n; ++i) $ for(intj =1; J < N; ++J)//enumerates the upper-left corner of a squarePanax Notoginseng for(intL =1; I+l<=n && j+l<=n; ++L)//enumerating the side length of a square - if(a[0][i][j+l]-a[0][I][J] = = L && a[0][i+l][j+l]-a[0][I+L][J] = =L the&& a[1][i+l][j]-a[1][I][J] = = L && a[1][i+l][j+l]-a[1][I][J+L] = =l) +cnt[l]++; A the if(Kase) printf ("\n**********************************\n\n"); +printf"problem #%d\n\n", ++Kase); - BOOLFlag =false; $ for(inti =1; I <= N; ++i)if(Cnt[i]) $ { -printf"%d square (s) of size%d\n", Cnt[i], i); -Flag =true; the } - if(!flag) puts ("No completed squares can be found.");Wuyi } the - return 0; Wu}

code June

UVa 201 Squares