1045. Favorite Color Stripe (30)

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:

65 2 3 1 5 612 2 2 4 1 5 5 6 3 1 1 5 6

Sample Output:

7

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相当于求最大公共子序列的长度，不过这里的最大公共子序列特别之处是数字可以重复（只要数字顺序符合第一个序列的顺序）。这里按照LCS算法的思路，用动态规划来做。用二维数组dp来保存当前状况的最大公共子序列长度（dp[i][j]表示第一个序列长度为i和第二个序列长度为j时的最大公共子序列长度），如果两个序列的当前位置数字一样则取dp[i][j-1]和dp[i-1][j]的最大值加上一来更新dp[i][j]，如果不一样则取dp[i][j-1]和dp[i-1][j]的最大值来更新dp[i][j]。最后输出dp[n][m]（n表示第一序列的长度，m表示第二序列的长度）。

代码：

#include <iostream>#include <vector>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;int main(){int N;cin>>N;int n;cin>>n;vector<int>a(n);for(int i=0;i<n;i++) cin>>a[i];int m;cin>>m;vector<int>b(m);for(int i=0;i<m;i++) cin>>b[i];vector<vector<int> >dp(n+1,vector<int>(m+1,0));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(a[i-1]==b[j-1]){dp[i][j]=max(dp[i][j-1]+1,dp[i-1][j]+1);}else {dp[i][j]=max(dp[i][j-1],dp[i-1][j]);}}}cout<<dp[n][m];}