PAT1045. Favorite Color Stripe

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:

7

代码

#include <cstdio>
#include <cstring>
bool have[202];
bool order[202][202];
int stripe[10002];
int dp[10002];
int max(int a,int b){return a > b ? a : b;}
int main(){
    int colorNum,orderNum,stripeNum;
    scanf("%d%d",&colorNum,&orderNum);
    int seq[202];
    for(int i = 0;i < orderNum;i++)scanf("%d",&seq[i]);
    scanf("%d",&stripeNum);
    for(int i = 1;i <= stripeNum;i++)scanf("%d",&stripe[i]);
    //order数组记录的是i在Eva喜欢的序列中是否在j的前面
    //如果i或j不是Eva喜欢的颜色order[i][j]为0，并且i可以等于j
    for(int i = 0;i < orderNum;i++)
    for(int j = i;j < orderNum;j++)
        order[seq[i]][seq[j]] = 1;
    //have数组记录的是 是否是Eva喜欢的颜色
    for(int i = 0;i < orderNum;i++)
        have[seq[i]] = 1;
    //处理方式和最长递增子序列相似
    for(int i = 1;i <= stripeNum;i++){
        int tmax = 0;
        if(have[stripe[i]])tmax = 1;
        for(int j = 1;j < i;j++)
            if(order[stripe[j]][stripe[i]]){
                 tmax = max(dp[j]+1,tmax);
            }
        dp[i] = tmax;
    }
    int ans = 0;
    for(int i = 1;i <= stripeNum;i++)
        ans = max(ans,dp[i]);
    printf("%d",ans);
}