過河問題

 N個人要從河的一側到另一次 一個船最多做2個人 這個船的速度受慢的人的影響。

輸入一堆數，爲每個人 的時間，怎麼規劃划船的總時間最少？

 

 

#include <iostream>
#include <list>
#include <algorithm>
using namespace std;
void input();
void sortlist();
void show();
void solution(int sel);
list<int> timelistA,timelistB;
int time;

void main()
{
    input();
    sortlist();
    solution(0);
    cout<<endl<<"總用時"<<time<<"秒"<<endl<<endl;
}

void solution(int sel)
{
    timelistA.sort();
    timelistB.sort();

    if (timelistA.size()==0)
    {
        return;
    }
    if(0==sel)  // 左->右
    {
        if (timelistA.size()>1)
        {
            list<int >::iterator a=timelistA.begin();
            list<int >::iterator b=timelistA.begin();b++;

        timelistB.push_back(*a);
        cout<<*a;
        timelistA.erase(a);
        cout<<" "<<*b<<"過去"<<endl;
        timelistB.push_back(*b);
        time+=*b;
        timelistA.erase(b);

        }
        else
        {
                cout<<timelistA.front()<<" "<<"過去"<<endl;
                timelistB.push_back(timelistA.front());
                time+=timelistA.front();
                timelistA.erase(timelistA.begin());

        }
        solution(1);
    }
    else if(1==sel) //右->左 回來
    {
    cout<<timelistB.front()<<"  回來"<<endl;
    timelistA.push_back(timelistB.front());
    time+=timelistB.front();
    timelistB.erase(timelistB.begin());

    solution(2);
    }
    else if (2==sel)//左->右
    {
        list<int >::iterator a=timelistA.end();
        a--;
        list<int >::iterator b=a;
        b--;
        timelistB.push_back(*a);
        timelistB.push_back(*b);
        cout<<*a<<" ";
        cout<<*b<<" 過去"<<endl;
            time+=*a;
        timelistA.erase(a);
        timelistA.erase(b);
        solution(3);
    }
    else if (3==sel)
    {

        cout<<timelistB.front()<<" 過來"<<endl;
            time+=timelistB.front();
            timelistA.push_back(timelistB.front());
        timelistB.erase(timelistB.begin());
        solution(0);
    }

}


void input()
{
    cout<<"input every one :"<<endl;

    while(1)
    {
        int a;
        cin>>a;
        timelistA.push_back(a);

        if (getchar()=='\n')
        {
            break;
        }
    }
    cout<<endl;
}

void sortlist()
{
    timelistA.sort();
}

void show()
{
    list<int>::iterator a=timelistA.begin();
    for(;a!=timelistA.end();++a)
        cout<<*a<<"\t";
}