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UVa 11583 Partitioning by Palindromes
Description
Problem H: Partitioning by Palindromes
We say a sequence of characters is a palindrome if it is the same written forwards and backwards. For example, ‘racecar‘ is a palindrome, but ‘fastcar‘ is not.
A partition of a sequence of characters is a list of one or more disjoint non-empty groups of consecutive characters whose concatenation yields the initial sequence. For example, (‘race‘, ‘car‘) is a partition of ‘racecar‘ into two groups.
Given a sequence of characters, we can always create a partition of these characters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groups needed for a given string such that every group is a palindrome?
For example:
- ‘racecar‘ is already a palindrome, therefore it can be partitioned into one group.
- ‘fastcar‘ does not contain any non-trivial palindromes, so it must be partitioned as (‘f‘, ‘a‘, ‘s‘, ‘t‘, ‘c‘, ‘a‘, ‘r‘).
- ‘aaadbccb‘ can be partitioned as (‘aaa‘, ‘d‘, ‘bccb‘).
Input begins with the number n of test cases. Each test case consists of a single line of between 1 and 1000 lowercase letters, with no whitespace within.
For each test case, output a line containing the minimum number of groups required to partition the input into groups of palindromes.
Sample Input
3 racecar fastcar aaadbccb
Sample Output
1 7 3
Kevin Waugh
题意：求出字符串的最小回文串个数。
思路：dp求回文串，最后求最小个数。
AC代码：
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.*;
public class Main{
public static void main(String[] args) throws IOException {
//Scanner scan=new Scanner(System.in);
StreamTokenizer st = new StreamTokenizer(new BufferedReader(
new InputStreamReader(System.in)));
st.nextToken();
int n=(int)st.nval;
//scan.nextLine();
for(int i=0;i<n;i++){
st.nextToken();
String s=(String)st.sval;
int len=s.length();
char a[]=new char[len];
a=s.toCharArray();
int dp[][]=new int[1010][1010];
for(int j=0;j<len;j++){
dp[j][j]=1;
dp[j+1][j]=1;
}
for(int j=len-1;j>=0;j--){
for(int k=j+1;k<len;k++){
dp[j][k]=Math.min(dp[j+1][k]+1, dp[j][k-1]+1);
if(a[j]==a[k]&&dp[j+1][k-1]==1){
dp[j][k]=Math.min(dp[j][k], 1);
}
}
}
for(int j=0;j<len;j++){
for(int k=0;k<j;k++){
dp[0][j]=Math.min(dp[0][j], dp[0][k]+dp[k+1][j]);
}
}
System.out.println(dp[0][len-1]);
}
}
}