Algorithm/Algorithm Problem

SWEA 4008 숫자만들기(DFS)

땅지원 2022. 10. 12. 17:21
 

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어떻게 배치하는게 최대인지 최소인지 모르니까 완탐 돌리는거 밖에 답이없음

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Solution_4008 {
    static int N,max_value,min_value;
    static int[] data, oper;


    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int T = Integer.parseInt(st.nextToken());

        for (int tc = 1; tc < T+1; tc++) {
            st = new StringTokenizer(br.readLine());

            N = Integer.parseInt(st.nextToken());

            oper = new int[4];
            data = new int[N];

            // {+ - * /}
            st = new StringTokenizer(br.readLine());
            for (int i = 0; i < 4; i++) {
                oper[i] = Integer.parseInt(st.nextToken());
            }

            st = new StringTokenizer(br.readLine());
            for (int i = 0; i < N; i++) {
                data[i] = Integer.parseInt(st.nextToken());
            }

            max_value = Integer.MIN_VALUE;
            min_value = Integer.MAX_VALUE;

            dfs(1,data[0]);

            System.out.println("#"+tc+" "+(max_value-min_value));



        }


    }


    public static void dfs(int depth, int sum){
        if (depth == N){
            max_value = Math.max(max_value, sum);
            min_value = Math.min(min_value, sum);

            return;
        }

        for (int i = 0; i < 4; i++) {
            if (oper[i] == 0)
                continue;

            oper[i]--;
            switch (i){
                case 0:
                    dfs(depth+1,sum+data[depth]);
                    break;
                case 1:
                    dfs(depth+1,sum-data[depth]);
                    break;
                case 2:
                    dfs(depth+1,sum*data[depth]);
                    break;
                case 3:
                    dfs(depth+1,sum/data[depth]);
                    break;
            }
            oper[i]++;


        }



    }

}