2023-06-13 BOJ 1932 정수 삼각형

문제 풀이 방식

• DP를 사용해서 풀면 된다.

문제 풀이 (Python)

import sys
r = sys.stdin.readline

n = int(r())
DP = [ list(map(int, r().rstrip().split())) for _ in range(n)]

step = 2

for i in range(1,n):
    for j in range(step):
        if j == 0:
            DP[i][j] = DP[i][j] + DP[i-1][j]
        elif i == j:
            DP[i][j] = DP[i][j] + DP[i-1][j-1]
        else:    
            DP[i][j] = DP[i][j] + max(DP[i-1][j-1], DP[i-1][j])
            
    step += 1
    
print(max(DP[n-1]))
            

문제 풀이(Java)

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

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());

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

        int step = 2;

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < step; j++) {
                if (j == 0) {
                    DP[i][j] = DP[i][j] + DP[i - 1][j];
                } else if (i == j) {
                    DP[i][j] = DP[i][j] + DP[i - 1][j - 1];
                } else {
                    DP[i][j] = DP[i][j] + Math.max(DP[i - 1][j - 1], DP[i - 1][j]);
                }
            }
            step += 1;
        }

        int maxSum = 0;
        for (int i = 0; i < n; i++) {
            maxSum = Math.max(maxSum, DP[n - 1][i]);
        }

        System.out.println(maxSum);
    }
}