專題測試三 貪心與動態規劃 H - Palindromic Subsequence
阿新 • • 發佈：2022-02-17
- 題目
A Subsequence is a sequence obtained by deleting zero or more characters in a string. A Palindrome is
a string which when read from left to right, reads same as when read from right to left. Given a string,
find the longest palindromic subsequence. If there are many answers to it, print the one that comes
lexicographically earliest.
Constraints
• Maximum length of string is 1000.
• Each string has characters ‘a’ to ‘z’ only.
Input
Input consists of several strings, each in a separate line. Input is terminated by EOF.
Output
For each line in the input, print the output in a single line.
Sample Input
aabbaabb
computer
abzla
samhita
Sample Output
aabbaa
c
aba
aha - 思路
尋找不連續的迴文字串，dp[i][j]表示從i到j的最長迴文串字元數，str表示字串，如果s[i]==s[j]，就在兩端加上這個字元，否則取dp[i+1][j]和dp[i][j-1]中字串最長、字典序最小的。 - 程式碼
#include<cstdio> #include<iostream> #include<string> #include<cstring> #include<algorithm> using namespace std; int dp[1050][1050]; string str[1050][1050]; char s[1050]; int n,i,j,k,p; int main() { while(scanf("%s",s+1)!=EOF) { n=strlen(s+1); for(i=1;i<=n;i++) { dp[i][i]=1; str[i][i]=s[i]; } for(k=1;k<=n;k++) for(i=1;i+k<=n;i++) { j=i+k; dp[i][j]=max(dp[i+1][j], dp[i][j-1]); if(s[i] == s[j]) { dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2); str[i][j]=s[i]+str[i+1][j-1]+s[j]; } else if(dp[i+1][j] == dp[i][j-1]) str[i][j]=min(str[i+1][j],str[i][j-1]); else if(dp[i][j] == dp[i+1][j]) str[i][j]=str[i+1][j]; else if(dp[i][j] == dp[i][j-1]) str[i][j]=str[i][j-1]; } cout<<str[1][n]<<endl; } }