AcWing 898. 数字三角形

题目链接

https://www.acwing.com/problem/content/900/

题目思路

这道题我们可以从最底层考虑，若要使当前路径之和最大，选择当前的所选值加左上/右上之和最大即可
因为有负数情况，所以要把边界设置好

题目代码

#include <iostream>
#include <algorithm>

using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 510;
int a[N][N], f[N][N];
int n;

int main()
{
    cin >> n;
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= i; j ++ )
            cin >> a[i][j];
            
    for(int i = 0; i <= n; i ++ )
        for(int j = 0; j <= i + 1; j ++ )   
            f[i][j] = -INF;
            
    f[1][1] = a[1][1];
    
    for(int i = 2; i <= n; i ++ )
        for(int j = 1; j <= i;j ++ )
            f[i][j] = max(f[i - 1][j - 1] + a[i][j], f[i - 1][j] + a[i][j]);
    
    int res = -INF;
    for(int i = 1; i <= n; i ++ )
        res = max(res, f[n][i]);
        
    cout << res << endl;
    return 0;
    
}

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来源： https://www.cnblogs.com/vacilie/p/16106386.html