1045 Favorite Color Stripe (30 point(s))

Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.

It is said that a normal human eye can distinguish about less than 200 different colors, so Eva’s favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.

Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva’s favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤200) which is the total number of colors involved (and hence the colors are numbered from 1 to N). Then the next line starts with a positive integer M (≤200) followed by M Eva’s favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L (≤104) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line a separated by a space.

Output Specification:

For each test case, simply print in a line the maximum length of Eva’s favorite stripe.

Sample Input:

6
5 2 3 1 5 6
12 2 2 4 1 5 5 6 3 1 1 5 6

Sample Output:

7

Code:

/*
因为喜欢的颜⾊是不重复的，把喜欢的颜⾊的序列依次存储到数组中，book[i] = j表示i颜⾊的下
标为j。先在输⼊的时候剔除不在喜欢的序列中的元素，然后把剩余的保存在数组a中。按照最⻓不下
降⼦序列的⽅式做，对于从前到后的每⼀个i，如果它前⾯的所有的j，⼀下⼦找到了⼀个j的下标
book[j]⽐book[i]⼩，此时就更新dp[i]使它 = max(dp[i], dp[j] + 1)；并且同时再每⼀次遍历完成⼀次j后
更新maxn的值为⻓度的最⼤值，最后输出maxn～
*/
#include <bits/stdc++.h>
using namespace std;

const int N = 1e4 + 10;

int book[210];
int a[N];
int dp[N];
int ans = 1;

int n, m, l;
int main()
{
    cin >> n;
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        int temp;
        cin >> temp;
        book[temp] = i;
    }

    cin >> l;
    for (int i = 1; i <= l; i++)
    {
        int temp;
        cin >> temp;
        if (book[temp] >= 1)
        {   //很重要
            a[ans++] = book[temp];
        }
    }

    ans--;
    
    for (int i = 1; i <= ans; i++)
    {
        dp[i] = 1;
        for (int j = 1; j < i; j++)
        {
            if (a[j] <= a[i])
            {
                dp[i] = max(dp[i], dp[j] + 1);
            }
        }
    }
    int res = 1;

    for (int i = 1; i <= ans; i++)
        res = max(res, dp[i]);

    cout << res;
}