UVa910 TV game-优快云博客

本文介绍了一种使用动态规划解决从起点到特定终点的路径计数问题的方法。通过定义dp(i,j)来表示第i步到达第j个点的路径数量，并利用状态转移方程实现高效求解。

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用dp(i,j)表示第i步到第j个点的计数，则状态转移方程为

dp(i,j) += dp(i - 1,k) ，当且仅当k到j存在边时

代码如下：

#include <iostream>
#include <fstream>
#include <vector>
#include <cstring>
#include <string>

class Solution
{
public:
	void init()
	{
		memset(move, 0x00, sizeof(move));
		memset(dp, 0x00, sizeof(dp));
		special.clear();
	}
	void addEdge(int s, int d1, int d2)
	{
		move[s][0] = d1; move[s][1] = d2;
	}

	void setn(int n)
	{
		this->n = n;
	}

	void setm(int m)
	{
		this->m = m;
	}

	void setSpecial(std::vector<int> v)
	{
		for (int spe : v)
		{
			special.push_back(spe);
		}
	}

	int solve()
	{
		dp[0][1] = 1;
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				for (int k = 1; k <= n; k++)
				{
					if (dp[i - 1][k] && move[k][0] == j)
					{
						dp[i][j] += dp[i - 1][k];
					}

					if (dp[i - 1][k] && move[k][1] == j)
					{
						dp[i][j] += dp[i - 1][k];
					}
				}
			}
		}

		int ans = 0;
		for (int v : special)
		{
			ans += dp[m][v];
		}

		return ans;

	}
private:
	static const int MAXV = 27;
	static const int MAX_MOVE = 31;

	int move[MAXV][2];
	std::vector<int> special;
	int n, m;
	int dp[MAX_MOVE][MAXV];
};

Solution solution;

int main()
{
#ifndef ONLINE_JUDGE
	std::ifstream fin("f:\\OJ\\uva_in.txt");
	std::streambuf* back = std::cin.rdbuf(fin.rdbuf());

	std::ofstream fout("f:\\OJ\\uva_out.txt");
	std::streambuf* outback = std::cout.rdbuf(fout.rdbuf());
#endif
	int n;
	while (std::cin >> n)
	{
		solution.init();
		solution.setn(n);
		std::vector<int> v;
		for (int i = 0; i < n; i++)
		{
			std::string s, d1, d2, spe;
			std::cin >> s >> d1 >> d2 >> spe;
			int from = s[0] - 'A' + 1;
			int to1 = d1[0] - 'A' + 1;
			int to2 = d2[0] - 'A' + 1;
			solution.addEdge(from, to1, to2);
			if (spe == "x")
			{
				v.push_back(from);
			}
		}
		solution.setSpecial(v);

		int m;
		std::cin >> m;
		solution.setm(m);
		int ans = solution.solve();
		std::cout << ans << std::endl;
	}
	
#ifndef ONLINE_JUDGE
	std::cin.rdbuf(back);
	std::cout.rdbuf(outback);
#endif
	return 0;
}