[BOJ] 1932번 : 정수 삼각형

1. problem : 

https://www.acmicpc.net/problem/1932

 

 

2. solution 1 :

#include <bits/stdc++.h>
using namespace std;
int n;
int d[505][505]; 
int trees[505][505]; 

int main(void) {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n; 
	int depth = 0;
	for (int i = 1; i <= n; i++) { // 층을 표현; 
		depth++; 
		for (int j = 1; j <= depth; j++) cin >> trees[i][j]; // 각 층의 몇번째 가지인지 표현;
	}
	d[1][1] = trees[1][1], d[2][1] = trees[2][1] + d[1][1], d[2][2] = trees[2][2] + d[1][1]; 
	
	for (int i = 3; i <= n; i++) {
		for (int j = 1; j <= i; j++) {
			if (j == 1) d[i][j] = d[i - 1][j] + trees[i][j];
			else if (j == i) d[i][j] = d[i - 1][j - 1] + trees[i][j];
			else d[i][j] = max(d[i - 1][j - 1], d[i - 1][j]) + trees[i][j];
		}
	}
	int maxVal = 0; 
	for (int i = 1; i <= n; i++) {
		if (maxVal < d[n][i]) maxVal = d[n][i];
	}
	cout << maxVal << '\n';
}

 

3. solution 2:

// Authored by : Hot6Mania
// Co-authored by : -
// http://boj.kr/b5c330a8f8684bca91511e8ea2a19a34
#include <bits/stdc++.h>
using namespace std;

int n;
int a[505][505], d[505][505];

int main(void){
  ios::sync_with_stdio(0);
  cin.tie(0);
  
  cin >> n;
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= i; ++j)
      cin >> a[i][j];
  d[1][1] = a[1][1];
  for(int i = 2; i <= n; ++i)
    for(int j = 1; j <= i; ++j)
      d[i][j] = max(d[i-1][j-1], d[i-1][j]) + a[i][j];
  cout << *max_element(d[n] + 1, d[n] + n + 1);
}

source code 출처 : https://github.com/encrypted-def/basic-algo-lecture/blob/master/0x10/solutions/1932.cpp

basic-algo-lecture/0x10/solutions/1932.cpp at master · encrypted-def/basic-algo-lecture