【线性dp】AcWing898.数字三角形

AcWing898.数字三角形

题解

自底向上

#include <iostream>

using namespace std;

const int N = 510;

int f[N][N];

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

自顶向下

#include <iostream>

using namespace std;

const int N = 510;

int f[N][N];

int main()
{
    int n;
    cin >> n;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= i; ++j)
            cin >> f[i][j];
    for(int i = 2; i <= n; ++i)
        for(int j = 1; j <= i; ++j)
    {
        if(j == 1) f[i][j] = f[i][j] + f[i-1][j];
        else if(j == i) f[i][j] = f[i][j] + f[i-1][j-1];
        else f[i][j] = max(f[i-1][j], f[i-1][j-1]) + f[i][j];
    }
    int res = -(1 << 30);
    for(int i = 1; i <= n; ++i) res = max(f[n][i], res);
    cout << res << endl;
    
    return 0;
}