poj1700过河问题（贪心）

Crossing River

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10871		Accepted: 4097

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

Source

POJ Monthly--2004.07.18

n个人过河，只有一条船，一条船最多装两个人，来回时间以最多那个人为标准，问过河最短时间

首先这个问题需要分类讨论假设有m个人，按时间大小排好序

m==1时，只有一个人过河 ,t+=a[1]

m==2时，两个人过河，直接t取大的,t+=a[2]

m==3时，三个人过河，先让最快的载最慢的过河，然后最快的回来，再在次快的过河,t+=a[3]+a[1]+a[2]

m>=4时，这里要用到贪心

无论m多大，我们只看4个人，最快 次最快 最慢 次最慢

一般我们能想到的就是第一种做法，让最快的载最慢的，然后最快的回来，再载次最慢的，然后最快的回来，这里我们不需要管次最快的，这里我们就减少了两个人了，因此m-=2,在去判断，m在哪个范围再进行求解就好了。

第二种，先给个样例 1 2 1000 1000 ，这个显然用第一种做法，是行不通的，1000+1+1000=2001，我们显然可以知道，先让最快和次快过河，然后最快回来，然后最慢和次慢过河，然后次快回来，2+1+1000=1003，贪心取时间最短，所以这就是最短时间的做法，这里我们还是没有把最快和次最快过河，因此还是减少两人，m-=2。

用上面两种做法，一直循环到边界就可以得出答案了。

#include 
#include 
#include 
#include 

using namespace std;
int a[1009];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,j,m,ans=0;
        scanf("%d",&n);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+1+n);
        m=n;
        while(m>0)
        {
            if(m>=4)
            {
                if(a[m]+a[m-1]+2*a[1]>a[2]+a[1]+a[m]+a[2])
                {
                    ans+=a[2]+a[1]+a[m]+a[2];
                }
                else
                {
                    ans+=a[m]+a[m-1]+2*a[1];
                }
                m-=2;
            }
            else if(m==3)
            {
                ans+=a[m]+a[1]+a[2];
                m=0;
            }
            else if(m==2)
            {
                ans+=a[m];
                m=0;
            }
            else if(m==1)
            {
                ans+=a[m];
                m=0;
            }
        }
        printf("%d\n",ans);
    }






}