Problem 18 : Maximum path sum I (5%)

Description

original

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3 7 4 2 4 6 8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

간략해석

위에서 아래로 내려갈때, 최대합을 구하시오.

Idea & Algorithm

naive idea

기본적인 DP문제입니다.

아래로 내려가면 각 칸에 도착했을 때, 최댓값을 DP배열에 저장하며 내려가면 됩니다.

advanced idea

위 설명을 식으로 전개하면 다음과 같습니다.

\[DP[i][j] = A[i][j] + max(DP[i-1][j-1], DP[i-1][j])\]

source code

#include <stdio.h>

int arr[100][100];

int main(){
    for(int i = 1 ; i <= 15 ; i++){
        for(int j = 1 ; j <= i; j++){
            scanf("%d",&arr[i][j]);
            arr[i][j] += arr[i-1][j-1]>arr[i-1][j]?arr[i-1][j-1]:arr[i-1][j];
        }
    }
    int max = 0;
    for(int i = 1 ; i <= 15 ; i++){
        max = arr[15][i] > max ? arr[15][i]:max;
    }
    printf("%d",max);
}

Problem 018