poj1700 Crossing River

Crossing River

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12585		Accepted: 4787

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

题意：n个人一条船，全部人要过河；船每次仅仅能载两个人。用时按最慢的算。求全部人过河的最小时间。

分析：两种情况；设最小的和第二小的为m1,m2，当前

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

int T,n;
int a[1010];
int sum;

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        CL(a, 0);
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        if(n == 1){printf("%d
",a[0]); continue;}
        sort(a, a+n);
        int m1 = a[0], m2 = a[1];
        sum = 0;
        for(int i=n-1; i>1; i-=2)
        {
            if(m2*2<m1+a[i-1])
            {
                if(i == 2) sum = sum+m1+m2+a[i];
                else sum = sum+m1+m2*2+a[i];
            }
            else
            {
                sum = sum+m1*2+a[i]+a[i-1];
                if(i == 2) sum -= a[0];
            }
        }
        if(n!=0&&n%2==0) sum += m2;
        printf("%d
",sum);
    }
    return 0;
}
</span>

最大的和第二大的为n1,n2；

1、m2+m2<m1+n2；m1和m2先过河，然后m1回来，n1和n2过河。m2回来；

2、m2+m2>m1+n2。m1和n1先过河。然后m1回来。m1和n2过河。m1回来；

另外还要注意奇数的情况。