UVa 201 Squares

题意：

给出这样一个图，求一共有多少个大小不同或位置不同的正方形。

分析：

这种题一看就有思路，最开始的想法就是枚举正方形的位置，需要二重循环，枚举边长一重循环，判断是否为正方形又需要一重循环，复杂度为O(n⁴)，对于n≤9来说，这个复杂度可以接受。

可以像预处理前缀和那样，用O(1)的时间判断是否为正方形，这样总的复杂度就优化到O(n³)。

这个方法转自这里

We can think that vertical or horizontal lines are edges between two adjecent point. After that 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 at 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 is 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 this just by subtracting values calculated above.

举个例子，a[i][j][0]表示在第i行上，从第一列到第j列水平边数，如果a[i][j+l][0] - a[i][j][0]，说明点(i, j)到(i, j+l)有一条长为l的水平线段。

我还被输入坑了，注意VH后面，哪个数代表行，哪个数代表列。

 1 #include <cstdio>

 2 #include <cstring>

 3 

 4 const int maxn = 10;

 5 bool G[2][maxn][maxn];

 6 int a[2][maxn][maxn], cnt[maxn];

 7 

 8 int main()

 9 {

10     //freopen("in.txt", "r", stdin);

11     int n, m, kase = 0;

12     while(scanf("%d", &n) == 1 && n)

13     {

14         memset(G, false, sizeof(G));

15         memset(a, 0, sizeof(a));

16         memset(cnt, 0, sizeof(cnt));

17         scanf("%d", &m);

18         getchar();

19         for(int k = 0; k < m; ++k)

20         {

21             char c;

22             int i, j;

23             scanf("%c %d %d", &c, &i, &j);

24             getchar();

25             if(c == 'H') G[0][i][j+1] = true;

26             else G[1][j+1][i] = true;

27         }

28         for(int i = 1; i <= n; ++i)

29             for(int j = 1; j <= n; ++j)

30             {

31                 a[0][i][j] = a[0][i][j-1] + G[0][i][j];

32                 a[1][i][j] = a[1][i-1][j] + G[1][i][j];

33             }

34 

35         for(int i = 1; i < n; ++i)

36             for(int j = 1; j < n; ++j)  //枚举正方形的左上角

37                 for(int l = 1; i+l<=n && j+l<=n; ++l)   //枚举正方形的边长

38                     if(a[0][i][j+l]-a[0][i][j] == l && a[0][i+l][j+l]-a[0][i+l][j] == l

39                        && a[1][i+l][j]-a[1][i][j] == l && a[1][i+l][j+l]-a[1][i][j+l] == l)

40                         cnt[l]++;

41 

42         if(kase) printf("\n**********************************\n\n");

43         printf("Problem #%d\n\n", ++kase);

44         bool flag = false;

45         for(int i = 1; i <= n; ++i) if(cnt[i])

46         {

47             printf("%d square (s) of size %d\n", cnt[i], i);

48             flag = true;

49         }

50         if(!flag) puts("No completed squares can be found.");

51     }

52 

53     return 0;

54 }

代码君