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 = 10;
bool G[2][maxn][maxn];
int a[2][maxn][maxn], cnt[maxn];

int main()
{
    //freopen("in.txt", "r", stdin);
    int n, m, kase = 0;
    while(scanf("%d", &n) == 1 && n)
    {
        memset(G, false, sizeof(G));
        memset(a, 0, sizeof(a));
        memset(cnt, 0, sizeof(cnt));
        scanf("%d", &m);
        getchar();
        for(int k = 0; k < m; ++k)
        {
            char c;
            int i, j;
            scanf("%c %d %d", &c, &i, &j);
            getchar();
            if(c == 'H') G[0][i][j+1] = true;
            else G[1][j+1][i] = true;
        }
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
            {
                a[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];
            }

        for(int i = 1; i < n; ++i)
            for(int j = 1; j < n; ++j)  //枚举正方形的左上角
                for(int l = 1; i+l<=n && j+l<=n; ++l)   //枚举正方形的边长
                    if(a[0][i][j+l]-a[0][i][j] == l && a[0][i+l][j+l]-a[0][i+l][j] == l
                       && a[1][i+l][j]-a[1][i][j] == l && a[1][i+l][j+l]-a[1][i][j+l] == l)
                        cnt[l]++;

        if(kase) printf("
**********************************

");
        printf("Problem #%d

", ++kase);
        bool flag = false;
        for(int i = 1; i <= n; ++i) if(cnt[i])
        {
            printf("%d square (s) of size %d
", cnt[i], i);
            flag = true;
        }
        if(!flag) puts("No completed squares can be found.");
    }

    return 0;
}