递推DP UVA 590 Always on the run

题目传送门

题意：题意难懂，就是一个小偷在m天内从城市1飞到城市n最小花费，输入的是每个城市飞到其他城市的航班。

分析：dp[i][j] 表示小偷第i天在城市j的最小花费。状态转移方程：dp[i][j] = min (dp[i-1][k] + cost[k][j][t%day]) t表示在t天时k飞往j的飞机的花费

收获：

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-29 14:07:43
* File Name     :UVA_590.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dp[1010][12];
int d[12][12];
int cost[12][12][32];

int main(void)    {
	int n, m, cas = 0;
	while (scanf ("%d%d", &n, &m) == 2)	{
		if (n == 0 && m == 0)	break;
		for (int i=1; i<=n; ++i)	{
			for (int j=1; j<=n; ++j)	{
				if (i != j)	{
					scanf ("%d", &d[i][j]);
					for (int k=0; k<d[i][j]; ++k)	{
						scanf ("%d", &cost[i][j][k]);
					}
				}
			}
		}

		memset (dp, INF, sizeof (dp));
		for (int i=2; i<=n; ++i)	{
			if (cost[1][i][0])	{
				dp[0][i] = cost[1][i][0];
			}
		}
		for (int i=1; i<m; ++i)	{
			for (int k=1; k<=n; ++k)	{
				for (int j=1; j<=n; ++j)	{
					if (j != k)	{
						int c = cost[j][k][i%d[j][k]];
						if (c)	dp[i][k] = min (dp[i][k], dp[i-1][j] + c);
					}
				}
			}
		}
		int ans = dp[m-1][n];

		printf("Scenario #%d\n", ++cas);
		if(ans != INF){
            printf("The best flight costs %d.\n\n", ans);
        }else{
            puts("No flight possible.\n");
        }
	}

    return 0;
}