POJ1700 Crossing River 过河问题

最新推荐文章于 2022-04-21 11:51:43 发布

原创最新推荐文章于 2022-04-21 11:51:43 发布 · 3.1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#each #integer #input #output #go

ACM&算法专栏收录该内容

21 篇文章

订阅专栏

本文探讨了一群人利用一艘仅能容纳两人的小船过河的问题，并提供了一个算法解决方案。该方案通过合理的人员配对策略，确保所有人能够以最短的时间全部过河。

一、题目信息

Crossing River

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8897		Accepted: 3323

Description

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

二、参考代码

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main()
{
    int N;
    cin >> N;
    while(N--)
    {
        vector<int> a;
        int n,m,sum = 0;
        cin >> n;
        for(int i = 1;i <= n;i++)
        {
            cin >> m;
            a.push_back(m);
        }
        sort(a.begin(),a.end());
        a.insert(a.begin(),0);
        while(1)
        {
            if(n == 2) 
            {
                sum += a[2];
                break;
            }
            else if(n == 1)
            {
                sum += a[1];
                break;
            }
            else if(n == 3)
            {
                sum +=a[1] + a[2] + a[3];
                break;
            }
            else
            {
                sum += min(2*a[2] + a[1] + a[n] ,a[n] + 2*a[1] + a[n-1]);
                a.erase(a.end()-1,a.end());
                n -= 2;
            }
        }
        cout << sum << endl;
    }
    return 0;
}