題干：

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?(≤10?4??) 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

題目大意：

Eva試著從給出的布條中選出她自己的彩色布條。她想要僅僅保留按照她最喜歡的順序排列的她喜歡的顏色，并且剪掉那些不喜歡的部分再將剩下的部分縫在一起來組成她最喜歡的彩色布條。所以她需要你的幫助來找出最好的結果。
注意結果可能不為1，但是你只需要告訴她最大的長度。舉個例子，給你一個 {2 2 4 1 5 5 6 3 1 1 5 6}.的彩色布條。如果Eva最喜歡的順序排列的最喜歡的顏色為{2 3 1 5 6},這樣她有4種可能的最佳方案 {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, 和{2 2 3 1 1 5 6}。（注意? 最喜歡的順序排列? 的每個數字都不重復）

解題報告：

因為最喜歡的順序排列的每個數字都不重復，也就是說我可以給這些數字重新定序。這樣就轉化為一個LIS問題了。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n,m,len; int a[222],b[MAX],dp[MAX],buf[MAX],top; int bk[222];//顏色出現的最后一個位置 int main() {memset(bk,-1,sizeof bk);//initcin>>n >> m;for(int i = 1; i<=m; i++) cin>>a[i],bk[a[i]] = i;cin>>len;for(int x,i = 1; i<=len; i++) {scanf("%d",&x);if(bk[x] != -1) {b[++top] = bk[x];}}for(int i = 1; i<=top; i++) {dp[i] = 1;for(int j = 1; j<i; j++) {if(b[i] >= b[j]) dp[i] = max(dp[i],dp[j]+1);}}printf("%d\n",*max_element(dp+1,dp+top+1));return 0 ; }

總結：

注意對于序列問題，可以用數字大小來表示前后關系。

?

還有另一種方法：看不太懂

https://blog.csdn.net/tjj1998/article/details/79951718

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<algorithm> #define INF 99999999 using namespace std; int N,M,L; int dp[210],v[210]; int main(){scanf("%d%d",&N,&M);int a;for(int i=0;i<M;i++){scanf("%d",&a);if(!v[a])v[a]=i+1;}scanf("%d",&L);for(int i=0;i<L;i++){scanf("%d",&a);if(v[a])for(int j=M;j>=v[a];j--){for(int k=v[a];k>=0;k--)dp[j]=max(dp[j],dp[k]+1);}}printf("%d\n",dp[M]);return 0; }

?

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