1045. Favorite Color Stripe (30)

Eva的最爱颜色条纹

最新推荐文章于 2024-12-11 16:19:42 发布

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最新推荐文章于 2024-12-11 16:19:42 发布

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Eva希望从一条包含多种颜色的长条纹中，通过裁剪掉不喜欢的颜色并缝合剩余部分来制作出她最喜爱的颜色组合条纹。本篇介绍了一个动态规划算法，用于找出最长的符合Eva喜好的颜色条纹。

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 (<=10000) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line are 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:

//动态规划题
//设dp[i] 为以优先级i结尾的符合要求的最长子串
//很容易推出状态转移方程 dp[i] = max(dp[1]...dp[i])+1  
//最后从dp[1]...dp[n]中取最大值即为结果。
#include "iostream"
#include "algorithm"
#include "cstring"
using namespace std;
int main() {
	int n, m;
	int prior[202];
	int dp[202];
	cin >> n;
	cin >> m;
	memset(prior, 0, sizeof(prior));
	for (int i = 0; i < m; i++) {
		int a;
		cin >> a;
		prior[a] = i+1; // 映射。
	}
	int p;
	int Max = 0;
	int MAX = 0;
	cin >> p;
	memset(dp, 0, sizeof(dp));
	for (int i = 0; i < p; i++) {
		int q;
		cin >> q;
		MAX = 0;
		for (int i = 1; i <= prior[q]; i++) {
			if (dp[i] > MAX) {
				MAX = dp[i];
			}
		}
		dp[prior[q]] = MAX + 1;
		for (int i = prior[q] + 1; i <= m; i++)
			if (dp[i] > Max)
				Max = dp[i];
		Max = max(MAX+1,Max);
	}
	cout << Max << endl;
	return 0;
}