Dynamic Programming (2)

1. Linear DP

1.1 Numerical triangle

 

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 510,INF = 1e9;

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

int main()
{
    scanf("%d",&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 <= 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]);

    printf("%d\n",res);

    return 0;
}

1.2 The longest increasing subsequence

 

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 1010;

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

int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++) scanf("%d",&a[i]);

    for(int i = 1;i <= n;i ++)
    {
        f[i] = 1; //终于a[i]一个数
        for(int j = 1;j < i;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]);

    printf("%d\n",res);

    return 0;
}

1.3 The Longest Common Subsequence Problem

#include<iostream>
#include<algorithm>

using namespace std;

const int N =1010;

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

int main()
{
    scanf("%d%d",&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[i]) f[i][j] = max(f[i][j],f[i - 1][j - 1] + 1);
        }

    printf("d\n",f[n][m]);

    return 0;
}