nyoj 716

River Crossing

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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).

输入

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)

输出

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

样例输入

2
2 10
3
5
5 10
3
4
6
100
1

样例输出

18
50

来源

第六届河南省程序设计大赛

/*
dp[i]表示人和i只羊一起过河所花费的最短时间
dp[i]=min(dp[i],dp[i-j]+m+dp[j]);
j:= 1~i-1;
*/
#include<stdio.h>
#include<algorithm>
using namespace std;
int dp[1005], t[1005];
int main()
{
    int T, n, m, i, j;
    scanf("%d",&T);
    while(T--)
    {
        int a;
        scanf("%d%d",&n, &m);
        t[0] = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&a);
            t[i] = t[i-1] + a;
        }
        for(i = 1; i <= n; i++)
        {
            dp[i] = t[i] + m;
            for(j = 1; j < i; j++)
                dp[i] = min(dp[i], dp[i-j] + dp[j] + m);
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}