【PAT甲级真题】- Favorite Color Stripe (30)

题目来源

注意点

颜色序列不用每个都出现，只要按顺序即可

题目描述

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}.

输入描述:

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.

输出描述:

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

输入例子:

复制代码

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

输出例子:

复制代码

思路简介

其实就是一道最长不降序列

把颜色的色号映射为输入的顺序

然后问题即转化为第二行的最长不降序列

注意题目当中的要求的颜色序列不用每个颜色都出现，只要按照顺序出现即可，可以跳过某个颜色

遇到的问题

以为所有颜色必须出现

代码

cpp 复制代码

#include<bits/stdc++.h>
using namespace std;

void solve(){
    int n;cin>>n;
    int m;cin>>m;
    vector<int>fav_index(n+1,0);// 颜色 -> 排名，不在喜欢列表中的颜色排名为0
    for(int i=1;i<=m;++i){
        int fav;
        cin>>fav;
        fav_index[fav]=i;
    }

    int l;cin>>l;
    vector<int>stripe;
    for(int i=0;i<l;++i){
        int color;
        cin>>color;
        if(!fav_index[color])continue;
        stripe.emplace_back(fav_index[color]);
    }


    int len=stripe.size();
    vector<int> dp(len,1);//dp(i)以i结尾的最长不降序列
    int ans=1;
    for(int i=0;i<len;++i){
        for(int j=0;j<i;++j){
            if(stripe[i]>=stripe[j]){
                dp[i]=max(dp[i],dp[j]+1);
            }
        }   
        ans=max(ans,dp[i]);
    }
    cout<<ans<<'\n';
}

int main() {
    ios::sync_with_stdio(false);cin.tie(nullptr);
    //fstream in("in.txt",ios::in);cin.rdbuf(in.rdbuf());

    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}