Crossing River

最新推荐文章于 2024-08-30 21:47:53 发布

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本文探讨了一群人使用一条最多可载两人且速度受慢者限制的小船过河的问题，通过模拟和策略选择，旨在找到使所有人过河所需时间最短的方法。

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A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has a different rowing speed; the speed of a couple is determined by the speed of the slower one. Your job is to determine a strategy that minimizes the time for these people to get across.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. Each case is preceded by a blank line. There won't be more than 1000 people and nobody takes more than 100 seconds to cross.

Output

For each test case, print a line containing the total number of seconds required for all the N people to cross the river.

Sample Input

1
4
1 2 5 10

Sample Output

n个人都想要过河，每一个人都有过河的时间，只有一条船，
最多能装2人，两个人驾驶一条船时，过河的时间
由两个人中的最长时间决定。问现在所有人过河最短需要多久。

当人数小于四时，可以直接模拟出来，当人数不小于四时，
将过河时间由小到大排序后，取a0,a1,ai-1,ai，这
四个人组成一组。有两种方案 1.a0先送ai过河然后回来
，再送ai-1，再回来。2. a0和a1过河，a0回去，然后
ai，ai-1过河，a1将船开回去。选择用时少的，
然后重复这个过程

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1010];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }

        sort(a,a+n);
        int ti=0;
        if(n==1){
            ti+=a[0];
        }else if(n==2){
            ti+=a[1];
        }else if(n==3){
            ti+=a[0]+a[1]+a[2];
        }else{
            int i;
            for(i=n-1;i>=3;i-=2){
                int t1=a[1]+a[0]+a[i]+a[1];
                int t2=a[0]+a[i]+a[0]+a[i-1];
                t1<t2?ti+=t1:ti+=t2;
            }
            if(i==1){
                ti+=a[1];
            }else{
                ti+=a[0]+a[1]+a[2];
            }
        }
        printf("%d\n",ti);
    }
}