ZOJ-1250

不难的DP，把每一天到各个城市的最小费用DP出来，然后明天就可以根据今天的情况继续递推最优解

#include<cstdio>
#include<vector>
#include<climits>

using namespace std;

int main()
{
	vector<int> V[10][10];
	int n, k, dp[1001][10], cs = 0;
	while (scanf("%d %d", &n, &k), n || k)
	{
		int day, cost;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				if (i != j)
				{
					V[i][j].clear();
					scanf("%d", &day);
					while (day--)
					{
						scanf("%d", &cost);
						V[i][j].push_back(cost);
					}
				}
		for (int i = 0; i <= k; i++)
			for (int j = 0; j < 10; j++)
				dp[i][j] = INT_MAX;
		dp[0][0] = 0;
		for (int t = 1; t <= k; t++)
			for (int i = 0; i < n; i++)
				for (int j = 0; j < n; j++)
					if (i != j && dp[t - 1][i] != INT_MAX)
					{
						int cc = V[i][j][(t - 1) % V[i][j].size()];
						if (cc && (dp[t - 1][i] + cc < dp[t][j]))
							dp[t][j] = dp[t - 1][i] + cc;
					}
		printf("Scenario #%d\n", ++cs);
		if (dp[k][n - 1] != INT_MAX)
			printf("The best flight costs %d.\n\n", dp[k][n - 1]);
		else
			puts("No flight possible.\n");
	}
	return 0;
}