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。

 #include <cstdio>
 #include <cstring>

 const int maxn = ;
 bool G[][maxn][maxn];
 int a[][maxn][maxn], cnt[maxn];

 int main()
 {
     //freopen("in.txt", "r", stdin);
     int n, m, kase = ;
     while(scanf("%d", &n) ==  && n)
     {
         memset(G, false, sizeof(G));
         memset(a, , sizeof(a));
         memset(cnt, , sizeof(cnt));
         scanf("%d", &m);
         getchar();
         for(int k = ; k < m; ++k)
         {
             char c;
             int i, j;
             scanf("%c %d %d", &c, &i, &j);
             getchar();
             if(c == 'H') G[][i][j+] = true;
             else G[][j+][i] = true;
         }
         for(int i = ; i <= n; ++i)
             for(int j = ; j <= n; ++j)
             {
                 a[][i][j] = a[][i][j-] + G[][i][j];
                 a[][i][j] = a[][i-][j] + G[][i][j];
             }

         for(int i = ; i < n; ++i)
             for(int j = ; j < n; ++j)  //枚举正方形的左上角
                 for(int l = ; i+l<=n && j+l<=n; ++l)   //枚举正方形的边长
                     if(a[][i][j+l]-a[][i][j] == l && a[][i+l][j+l]-a[][i+l][j] == l
                        && a[][i+l][j]-a[][i][j] == l && a[][i+l][j+l]-a[][i][j+l] == l)
                         cnt[l]++;

         if(kase) printf("\n**********************************\n\n");
         printf("Problem #%d\n\n", ++kase);
         bool flag = false;
         for(int i = ; i <= n; ++i) if(cnt[i])
         {
             printf("%d square (s) of size %d\n", cnt[i], i);
             flag = true;
         }
         if(!flag) puts("No completed squares can be found.");
     }

     return ;
 }

代码君