Always on the run - UVa 590 dp

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dp动态规划和递推同时被 2 个专栏收录

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本文探讨了一位艺术窃贼如何通过优化路径选择和航班预订来实现最小化逃逸成本的问题。面对每日航班价格周期性波动的挑战，作者运用动态规划算法，帮助窃贼在有限天数内从巴黎出发，最终抵达亚特兰大，同时尽可能减少旅行支出。

Always on the run

Screeching tires. Searching lights. Wailing sirens. Police cars everywhere. Trisha Quickfinger did it again! Stealing the `Mona Lisa' had been more difficult than planned, but being the world's best art thief means expecting the unexpected. So here she is, the wrapped frame tucked firmly under her arm, running to catch the northbound metro to Charles-de-Gaulle airport.

But even more important than actually stealing the painting is to shake off the police that will soon be following her. Trisha's plan is simple: for several days she will be flying from one city to another, making one flight per day. When she is reasonably sure that the police has lost her trail, she will fly to Atlanta and meet her `customer' (known only as Mr. P.) to deliver the painting.

Her plan is complicated by the fact that nowadays, even when you are stealing expensive art, you have to watch your spending budget. Trisha therefore wants to spend the least money possible on her escape flights. This is not easy, since airlines prices and flight availability vary from day to day. The price and availability of an airline connection depends on the two cities involved and the day of travel. Every pair of cities has a `flight schedule' which repeats every few days. The length of the period may be different for each pair of cities and for each direction.

Although Trisha is a good at stealing paintings, she easily gets confused when booking airline flights. This is where you come in.

Input

The input file contains the descriptions of several scenarios in which Trisha tries to escape. Every description starts with a line containing two integers n and k. n is the number of cities through which Trisha's escape may take her, and k is the number of flights she will take. The cities are numbered $1, 2, \dots, n$ , where 1 is Paris, her starting point, and n is Atlanta, her final destination. The numbers will satisfy $2 \leŸ n \leŸ 10$ and $1 \leŸ k \leŸ 1000$ .

Next you are given n(n - 1) flight schedules, one per line, describing the connection between every possible pair of cities. The first n - 1 flight schedules correspond to the flights from city 1 to all other cities ( $2, 3, \dots, n$ ), the next n - 1 lines to those from city 2 to all others ( $1, 3, 4, \dots, n$ ), and so on.

The description of the flight schedule itself starts with an integer d, the length of the period in days, with $1 \leŸ d \leŸ 30$ . Following this are d non-negative integers, representing the cost of the flight between the two cities on days $1, 2, \dots, d$ . A cost of 0 means that there is no flight between the two cities on that day.

So, for example, the flight schedule ``3 75 0 80'' means that on the first day the flight costs 75, on the second day there is no flight, on the third day it costs 80, and then the cycle repeats: on the fourth day the flight costs 75, there is no flight on the fifth day, etc.

The input is terminated by a scenario having n = k = 0.

Output

For each scenario in the input, first output the number of the scenario, as shown in the sample output. If it is possible for Trisha to travel k days, starting in city 1, each day flying to a different city than the day before, and finally (after k days) arriving in city n, then print ``The best flight costs x.'', where x is the least amount that the k flights can cost.

If it is not possible to travel in such a way, print ``No flight possible.''.

Print a blank line after each scenario.

Sample Input

3 6
2 130 150
3 75 0 80
7 120 110 0 100 110 120 0
4 60 70 60 50
3 0 135 140
2 70 80
2 3
2 0 70
1 80
0 0

Sample Output

Scenario #1
The best flight costs 460.

Scenario #2
No flight possible.

题意：一个人每天都要换一个城市，每天从一个城市到另一个城市的价钱会有周期性的变化，问你在m天后到达n城市最少的花费是多少。

思路：dp[i][j]表示在第i天当前在j点的最少花费，然后由j向i+1天递推。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[1010][15],cost[15][15][35];
int main()
{ int t=0,n,m,i,j,k,money,p;
  while(~scanf("%d%d",&n,&m) && n+m)
  { for(i=1;i<=n;i++)
     for(j=1;j<=n;j++)
     { if(i==j)
        continue;
       scanf("%d",&cost[i][j][0]);
       for(k=1;k<=cost[i][j][0];k++)
        scanf("%d",&cost[i][j][k]);
     }
    memset(dp,-1,sizeof(dp));
    dp[0][1]=0;
    for(i=0;i<m;i++)
     for(j=1;j<=n;j++)
     { if(dp[i][j]>=0)
        for(k=1;k<=n;k++)
        { if(j==k)
           continue;
          p=(i+1)%cost[j][k][0];
          if(p==0)
           p=cost[j][k][0];
          money=cost[j][k][p];
          if(money==0)
           continue;
          if(dp[i+1][k]==-1)
           dp[i+1][k]=dp[i][j]+money;
          else
           dp[i+1][k]=min(dp[i+1][k],dp[i][j]+money);
        }
     }
    printf("Scenario #%d\n",++t);
    if(dp[m][n]==-1)
     printf("No flight possible.\n\n");
    else
     printf("The best flight costs %d.\n\n",dp[m][n]);
  }
}