英文題目：

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

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中文描述：就是有一條河，只有一條船，有n個人，每個人有自己的過河時間，一條船隻能載兩個人，過河時間算兩個人中最大時間那個。算出最少過河時間。

個人理解：首先我做這道題題目時候，這個題目使用貪心策略；
首先我們來分析四個人的情況，因為3、2、1個人的時候很好辦；首先有a（最快）、b（次快）、c（次慢）、d（最慢），分兩個策略來解決問題。
1.a和b過河，a回來（保證時間最少），c和d過河，b回去（保證時間最少），保證下一次運輸時候，a和b來作為來回運輸的人，保證時間最少。
2.a和d過河，a回來，a和c過去，a回來。
所以我們只要在n>3個人的時候來比較這兩個策略那一個策略時間比較少，然後n-2；直到人數剩下3人一下（包括3個人）。

我們來程式碼實現：

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<math.h>
 4 #include<vector>
 5 #include<iterator>
 6 #include<string>
 7 #include<string.h>
 8 #include<ctype.h>
 9 #include<map>
10 #include<stack>
11 #include<queue>
12 #include<iostream>
13 #include<time.h>
14 
15 using namespace std;
16 
17 #define rep(i ,a, b) for(int i = a; i <= b; i++)
18 #define per(i, a, b) for(int i = a; i <= b; i--)
19 int a[1005];
20 
21 int main()
22 {
23     int t, n, sum;//最簡單的基礎貪心問題；
24     scanf("%d", &t);//過河有兩種策略；
25     while(t--)
26     {
27         n = 0;
28         sum = 0;
29         scanf("%d", &n);
30         for(int i = 0; i < n; i++)
31         {
32             scanf("%d", &a[i]);
33         }
34         sort(a, a+n);
35         while(n > 3)
36         {
37             sum = min(sum+a[1]+a[0]+a[n-1]+a[1],sum+a[n-1]+a[0]+a[n-2]+a[0]);
38             n-=2;//保證人數過去時時間最大的兩個人。
39         }
40         if(n == 3)
41         {
42             sum += a[0]+a[1]+a[2];
43         }
44         if(n==2)
45         {
46             sum +=a[1];
47         }
48         if(n==1)
49         {
50             sum += a[0];
51         }
52         printf("%d\n", sum);
53     }
54     return 0;
55 }

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