poj1923 Fourier's Lines

思路：

记忆化搜索。

n条直线的交点方案数 
=（n-r）条平行线与r条直线交叉的交点数+r条直线本身的交点方案 
=（n-r）*r+r条直线之间本身的交点方案数（0<r<=n）

于是可以枚举r，递归来计算。

实现：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 int dp[][];
 int dfs(int n, int m)
 {
     if (m <  || m > n * (n - ) >> ) return ;
     if (!m) return ;
     if (dp[n][m] != -) return dp[n][m];
     for (int i = ; i <= n; i++)
     {
         if (dfs(n - i, m - i * (n - i)))
             return dp[n][m] = ;
     }
     return dp[n][m] = ;
 }

 int main()
 {
     int kase = , n, m;
     memset(dp, -, sizeof(dp));
     while (cin >> n >> m, n + m)
     {
         if (dfs(n, m))
             cout << "Case " << kase++ << ": " << n << " lines with exactly " << m << " crossings can cut the plane into " << n + m +  << " pieces at most." << endl;
         else
             cout << "Case " << kase++ << ": " << n << " lines cannot make exactly " << m << " crossings." << endl;
     }
     return ;
 }