NYOJ716 River Crossing（第六届河南省程序设计大赛）

 

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

省赛的时候没时间做了，我一直在调其他题，这题是队友做的，没做出来，很可惜。比赛时候因为编译器问题大部分时间都在调试，耽误了不少时间，这题应该做出来的，真遗憾。

#include <iostream>
#include <cstdio>

using namespace std;

const int inf = 0x3fffffff;
int m, n;
int t[1005], min_time[1005];

int dp(int left)  //表示把数量为left的羊送到对岸，人又回来所需的最少时间
{
    if(left <= 0)
        return 0;
    if(min_time[left] != inf)
        return min_time[left];
    int ans = inf;
    for(int i = 1; i <= left; ++i)
    {
        if(dp(left-i) + t[i] + m < ans)
            ans = dp(left-i) + t[i] + m;
    }
    return min_time[left] = ans;
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &t[i]);
            if(i == 1)
                t[i] += m;
            t[i] += t[i-1];
            min_time[i] = inf;
        }
        printf("%d\n", dp(n) - m);
    }
    return 0;
}