【动态规划】—— 线性DP

#include 
#include 
using namespace std;

const int N = 510, INF = 1e9;

int n;
int a[N][N];
int f[N][N];

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

朴素做法的DP公式： 

#include 
#include 
using namespace std;

const int N = 1010;

int n;
int a[N], f[N];

int main()
{
    cin >> n;
    for(int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
    
    for(int i = 1; i <= n; i ++ )
    {
        f[i] = 1; // 只有a[i]一个数的情况
        for(int j = 1; j <= i - 1; j ++ )
            if(a[j] < a[i])
                f[i] = max(f[i], f[j] + 1);
    }
    
    int res = 0;
    for(int i = 1; i <= n; i ++ ) res = max(res, f[i]);
    
    cout << res << endl;
    
    return 0;
}

#include 
#include 
using namespace std;

const int N = 1010;

int n;
int a[N], f[N];
int g[N];

int main()
{
    cin >> n;
    for(int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);
    
    for(int i = 1; i <= n; i ++ )
    {
        f[i] = 1; // 只有a[i]一个数的情况
        for(int j = 1; j <= i - 1; j ++ )
            if(a[j] < a[i])
                if(f[i] < f[j] + 1)
                {
                    f[i] = f[j] + 1;
                    g[i] = j;
                }
    }
    
    int k = 1;
    for(int i = 1; i <= n; i ++ )
        if(f[k] < f[i])
            k = i;
    
    for(int i = 0, len = f[k]; i < len; i ++ )
    {
        cout << a[k] << ' ';
        k = g[k];
    }
    
    return 0;
}

#include 
#include 

using namespace std;

const int N = 1010;

int n, m;
char a[N], b[N];
int f[N][N];

int main()
{
    cin >> n >> m;
    scanf("%s%s",a + 1, b + 1);
    
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= m; j ++ )
        {
            f[i][j] = max(f[i - 1][j], f[i][j - 1]);
            if(a[i] == b[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
        }
    cout << f[n][m] << endl;
    return 0;
}