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[105][5005];
int dfs(int n, int m)
{
    if (m < 0 || m > n * (n - 1) >> 1) return 0;
    if (!m) return 1;
    if (dp[n][m] != -1) return dp[n][m];
    for (int i = 1; i <= n; i++)
    {
        if (dfs(n - i, m - i * (n - i)))
            return dp[n][m] = 1;
    }
    return dp[n][m] = 0;
}

int main()
{
    int kase = 1, n, m;
    memset(dp, -1, 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 + 1 << " pieces at most." << endl;
        else
            cout << "Case " << kase++ << ": " << n << " lines cannot make exactly " << m << " crossings." << endl;
    }
    return 0;
}