渡河问题(Crossing River)-贪心策略

Posted 2021-09-22 nuist__NJUPT

Crossing River

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

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void greedy(int N, int [] speeds){
        int left = N ; //还未渡河的人数
        int times = 0 ; //初始化渡河所需的总时间
        while(left > 0){
            if(left == 1){
                times += speeds[0] ;
                break ;
            }else if(left == 2){
                times += speeds[1] ;
                break ;
            }else if(left == 3){
                times += speeds[0] + speeds[1] + speeds[2] ;
                break ;
            }else{//两种渡河策略
                //快慢分组法：1，2渡河，1回来，3，4渡河，2回来
                int s1 = speeds[1] + speeds[0] + speeds[left-1] + speeds[1] ;
                //快带慢法：1，4渡河，1回来，1，3渡河，1回来
                int s2 = speeds[left-1] + speeds[left-2] + 2 * speeds[0] ;
                times += Math.min(s1, s2) ;
                left -= 2 ; //所需渡河人数减少2
            }
        }
        System.out.println(times) ;
    }
    public static void main(String[] args){
        Scanner input = new Scanner(System.in) ;
        int T = input.nextInt() ;
        for(int i=0; i<T; i++){
            int N = input.nextInt() ;
            int [] speeds = new int [N] ;
            for(int j=0; j<N; j++){
                speeds[j] = input.nextInt() ;
            }
            Arrays.sort(speeds) ;
            greedy(N, speeds) ;
        }

    }
}