快速渡河（贪心）
/*
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
*/
package _90贪心;
import java.util.Arrays;
import java.util.Scanner;
/*1是组数据
* 4 四个入 1和2过去 2秒
* 1 2 5 10 四个人的速度 1回来 花1秒 5和10 过去花10秒 2回来花两秒 1和2过去花2秒
* 17 输出最优解；
*/
//a b c d == a+3b+d //分组
//a b c d ==2a+b+c+d 快的带
public class b快速渡河 {
public static void main(String[] args) {
Scanner in=new Scanner(System.in) ;
int T =in.nextInt();
for(int i=0;i<T;i++) {
int n=in.nextInt();
int[]speed=new int[n];//可以放在一个循环里
for(int j=0;j<n;j++) {
speed[j]=in.nextInt();
}
Arrays.sort(speed);
f(n,speed);
}
}
private static void f(int n, int[] speed) {
int left=n;//有多少人
int ans=0;//记时间
while(left>0) {
if(left==1) {
ans+=speed[0];
break;
}else if(left==2) {
ans+=speed[1];
break;
}else if(left==3) {
ans+=speed[2]+speed[0]+speed[1];
break;
}else {
//1，2出发，1返回，最后两名出发，2返回
int s1=speed[1]+speed[0]+speed[left-1]+speed[1];//2b+a+d
//1，3出发，1返回，1，4出发，1返回，1，2过河
int s2=speed[left-1]+speed[left-2]+2*speed[0];//c+d+2a
ans +=Math.min(s1, s2);
left-=2;;//左侧是渡河的起点，left代表左侧的剩余人数
}
}
System.out.println(ans);
}
}