1045 Favorite Color Stripe (30分)(简单dp)

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 (≤) 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 (≤) followed by M Eva's favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L (≤) 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

题目分析:按照要求要截取袋子 但每次截取都必须按照给定的顺序来截取 对于每一种颜色 利用先后顺序为它编号 读取过程中去除掉未被编号的袋子 对余下的袋子进行动态规划 
对于每个dp[i]来说 从j到i重新比较 dp[i]与dp[j]+1 最后找到最大值

#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
int color[201];
int dp[10001];
int Array[10001];
int main()
{
    int N, M;
    cin >> N >> M;
    int col;
    for (int i = 1; i <= M; i++)
    {
        cin >> col;
        color[col] = i;
    }
    int K, num = 0;;
    cin >> K;
    for (int i = 0; i < K; i++)
    {
        cin >> col;
        if (color[col])
            Array[num++] = color[col];
    }
    int maxn = 0;
    for (int i = 0; i<num; i++)
    {
        dp[i] = 1;
        for (int j = 0; j < i; j++)
            if (Array[j] <= Array[i])
                dp[i] = max(dp[i], dp[j] + 1);
        maxn = max(maxn, dp[i]);
    }
    cout << maxn;
}