【nyoj716】River Crossing

最新推荐文章于 2021-05-08 19:48:45 发布

原创最新推荐文章于 2021-05-08 19:48:45 发布 · 281 阅读

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本文探讨了一个经典的运输问题：一个人如何将n头羊通过一条河，同时考虑到每增加一头羊过河时间会相应增加的情况，寻求将所有羊都安全送达对岸所需的最短时间。

Description

　Afandi is herding N sheep across the expanses of grassland when he finds himself blocked by a river. A single raft is available for transportation.

Afandi knows that he must ride on the raft for all crossings, but adding sheep to the raft makes it traverse the river more slowly.

When Afandi is on the raft alone, it can cross the river in M minutes When the i sheep are added, it takes Mi minutes longer to cross the river than with i-1 sheep (i.e., total M+M1 minutes with one sheep, M+M1+M2 with two, etc.).

Determine the minimum time it takes for Afandi to get all of the sheep across the river (including time returning to get more sheep).

Input

On the first line of the input is a single positive integer k, telling the number of test cases to follow. 1 ≤ k ≤ 5 Each case contains:

* Line 1: one space-separated integers: N and M (1 ≤ N ≤ 1000 , 1≤ M ≤ 500).

* Lines 2..N+1: Line i+1 contains a single integer: Mi (1 ≤ Mi ≤ 1000)

Output

For each test case, output a line with the minimum time it takes for Afandi to get all of the sheep across the river.

Sample Input

2 10

5 10

100

Sample Output

题意：一个人带着n头羊过河，自己过河时间是m，带一头羊是m+m1，带两头羊时间m+m1+m2，...求把所有羊运送过河所用的最短时间。

思路：dp[i]为运送i只羊所用的最短时间，dp[i]=min(dp[i],dp[i-x]+dp[x]+m),x从1~i-1

代码：

#include<bits/stdc++.h>
using namespace std;
int dp[1005];
const int inf=0x3f3f3f3f;
int main()
{
    int n,m,i,x,t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d",&n,&m);
        dp[0]=m;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&x);
            dp[i]=dp[i-1]+x;
        }
        for(i=1; i<=n; i++)
            for(x=1; x<=i-1; x++)
                dp[i]=min(dp[i],dp[i-x]+dp[x]+m);
        printf("%d\n",dp[n]);
    }
    return 0;
}