POJ 1700----Crossing River（过桥问题）

最新推荐文章于 2021-02-08 10:56:52 发布

原创最新推荐文章于 2021-02-08 10:56:52 发布 · 519 阅读

0 ·

CC 4.0 BY-SA版权

博弈论专栏收录该内容

2 篇文章

订阅专栏

本文探讨了一群人利用一艘只能容纳两人的船渡河的问题，并提供了一个算法解决方案。该算法通过合理安排人员往返渡河的方式，以最小化所有人完成渡河所需的总时间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

17

#include <stdio.h>  
#include <algorithm>      
using namespace std;      
#define N 1010
int a[N];  
int main()  
{ 
    int T,i,n,s;
    scanf("%d", &T);  
    while(T--)  
    {  
        scanf("%d", &n);  
        for(i=0;i<n;i++)  scanf("%d", &a[i]);  
        sort(a,a+n);  
        s=0;  
        while(n)  
        {  
            if(n==1)  
            {  
                s+=a[0];  
                break;  
            }  
            else if(n==2)  
            {  
                s+=a[1];  
                break;  
            }  
            else if(n==3)  
            {  
                s+=a[0]+a[1]+a[2];  
                break;  
            }  
            else  
            {  //模式1:A将Z送过桥，然后返回，再把Y送过桥，再返回；
	      //模式2:A和B先过桥，然后A返回，Y和Z过桥，然后B返回
                if(2*a[1]>a[0]+a[n-2]) s+=2*a[0]+a[n-1]+a[n-2];  //当2b>a+y时，使用模式一
                else s+=a[0]+a[n-1]+a[1]*2;  //使用模式2 
            }  
            n-=2;     
        }   
		printf("%d\n", s);    
    } 
	return 0;
}