poj1700 Crossing River

jlqwer 发表于 代码 分类,标签: 热度:112°
  • 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 <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
    int t,n,x[1005];
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(int i=0;i<n;i++)
            cin>>x[i];
        sort(x,x+n);
        int index=n,sum=0;    //index 还有多少人没过河 sum 已用的秒数
        while(index)    //只剩3个人时直接推结果
        {
            if(index==1)
            {
                sum+=x[0];
                break;
            }
            else if(index==2)
            {
                sum+=x[1];
                break;
            }
            else if(index==3)
            {
                sum+=x[1]+x[2]+x[0];
                break;
            }
            else
            {
                sum+=min(2*x[1]+x[index-1]+x[0],2*x[0]+x[index-1]+x[index-2]);
                //1、 a和b出发,a回来,c和d出发,b回来
                //2、 a和c出发,a回来,a和d出发,a回来
                index-=2;
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}


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