PAT 1034Favorite Color Stripe (30)(官网测试通过)

题目

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:65 2 3 1 5 612 2 2 4 1 5 5 6 3 1 1 5 6输出例子:7

思路

• 1.想了好久真的都想不到要用动态规划画一个表出来，其中like为行，表示题目中的喜欢的序列，s为原来的丝带的颜色序列。表p如下：

LIKE\S 2 2 4 1 5 5 6 3 1 1 5 6 2 1 2 2 2 2 2 2 2 2 2 2 2 23 1 2 2 2 2 2 2 3 3 3 3 3 231 1 2 2 3 3 3 3 3 4 5 5 5 2315 1 2 2 3 4 5 5 5 5 5 5 6 23156 1 2 2 3 4 5 6 6 6 6 6 7

• 2.对比第二行23和第一行2的区别，可以发现，假如没有3，那么第二排就跟第一排是一样的，来了个三有了什么区别呢？来了个3，且如果s[j]==like[i]的话（即s数列后面遇到了一个三）则意味着要在前一行的同一列的基础上加一，否则不变，数学语言为：if(s[j]==like[i]){p[i][j]=p[i-1][j]+1}，如表中加粗的3，则为第二行与第一行的区别所在。同时，后面如果继续还来3的话还要在同一行前面一个数的基础上加一，再与前面的如果一直没遇到3就与前一行一致，联合起来就变为取p[i-1][j],p[i][j]的最大值。即：else{s[i][j]=max(s[i][j-1],s[i-1][j])}。
• 3.都知道动态规划是要找一个表，想不到是要找这样子的表。

代码提示

• 1.我是先将第一行搞定，然后把第一列搞定，然后再根据这俩个为基础吧上述表填完的。
• 2.不知道为什么要构建这个表的话去看我的这篇博客简谈动态规划

代码

#include<iostream>#include<vector>#include<string>#include<iomanip>#include<sstream>#include<algorithm>using namespace std;int p[10005][205];int main(){    int n; cin >> n;    int m;    cin >> m;    vector<int> like(m);    for (int i = 0; i < m; i++)    {        cin >> like[i];    }    int l; cin >> l; vector<int> s(l);    int num = 0;    for (int i = 0; i < l; i++)    {        cin >> s[i];        //填好第一行        if (s[i]==like[0])        {            num++;        }        p[0][i] = num;    }    for (int i = 1; i < m; i++)    {        for (int j = 0; j < l; j++)        {        //填第一列            if (j==0)            {                if (like[i] == s[0])                {                    p[i][0] = p[i - 1][0] + 1;                }                else                    p[i][0] = p[i - 1][0];            }            else            {            //根据第一行和第一列吧剩下的填完                if (s[j] == like[i])                {                    p[i][j] = p[i][j - 1] + 1;                }                else                {                    p[i][j] = max(p[i][j - 1], p[i - 1][j]);                }            }        }    }    cout << p[m - 1][l - 1] << endl;    return 0;}