PAT-A1045 Favorite Color Stripe 题目内容及题解

最新推荐文章于 2024-02-20 19:49:20 发布

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最新推荐文章于 2024-02-20 19:49:20 发布

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该博客主要解析PAT(A1045)编程题，Eva试图从给定颜色条纹中按偏好保留最喜欢的颜色。题目要求找出保留最多数量的最喜欢颜色的条纹，输出最大长度。博客内容包括输入输出规格、题目解释、解题思路、代码实现和运行结果。

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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 (≤104) 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

题目大意

伊娃有一个长布条，但是她希望只以她喜欢的顺序保留她最喜欢的颜色，剪下那些不想要的部分，把剩下的部分缝在一起，形成她最喜欢的颜色条纹。原始布条可能很长，而Eva希望保留最长的最喜欢的布条。所以她需要你的帮助，为她找到最好的结果。结果可能不唯一，但是只需要返回最大的长度。

解题思路

读入数据并初始化动态规划边界；
采用动态规划方式确定结果；
输出结果并返回零值。

代码

#include<stdio.h>
#define maxc 210
#define maxn 10010

int Max(int a,int b){
    return a>b?a:b;
}

int A[maxc],B[maxn],dp[maxc][maxn];

int main(){
    int N,M,L;
    int i,j,MAX;
    scanf("%d%d",&N,&M);
    for(i=1;i<=M;i++){
        scanf("%d",&A[i]);
    }
    scanf("%d",&L);
    for(i=1;i<=L;i++){
        scanf("%d",&B[i]);
    }
    //边界
    for(i=0;i<=M;i++){
        dp[i][0]=0;
    }
    for(j=0;j<=L;j++){
        dp[0][j]=0;
    }
    //状态转移方程
    for(i=1;i<=M;i++){
        for(j=1;j<=L;j++){
            MAX=Max(dp[i-1][j],dp[i][j-1]);
            if(A[i]==B[j]){
                dp[i][j]=MAX+1;
            }else{
                dp[i][j]=MAX;
            }
        }
    }
    printf("%d\n",dp[M][L]);
    return 0;
}