UVa11584 - Partitioning by Palindromes(动态规划)

We say a sequence of characters is a palindrome if it is the same writtenforwards and backwards. For example, 'racecar' is a palindrome, but 'fastcar' is not.

A partition of a sequence of characters is a list ofone 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 thesecharacters such that each group in the partition is a palindrome! Given this observation it is natural to ask: what is the minimum number of groupsneeded 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 ofbetween 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

用dp(i)表示 到第i个字符时最小的回文数，状态转移方程为：
dp(i) = min{dp(j - 1) + 1} 1 <= j <= i ,如果(j,i)之间的字符是回文字符串

import java.io.FileInputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.PrintWriter;
import java.io.OutputStreamWriter;

public class Main implements Runnable{
private static final boolean DEBUG = false;
private static final int N = 1005;
private BufferedReader cin;
private PrintWriter cout;
private StreamTokenizer tokenizer;
private String s;
private int[] dp;

private void init()
{
try {
if (DEBUG) {
cin = new BufferedReader(new InputStreamReader(
new FileInputStream("e:\\uva_in.txt")));
} else {
cin = new BufferedReader(new InputStreamReader(System.in));
}

tokenizer = new StreamTokenizer(cin);
cout = new PrintWriter(new OutputStreamWriter(System.out));
} catch (Exception e) {
e.printStackTrace();
}
}

private String next()
{
try {
tokenizer.nextToken();
if (tokenizer.ttype == StreamTokenizer.TT_EOF) return null;
else if (tokenizer.ttype == StreamTokenizer.TT_NUMBER) {
return String.valueOf((int)tokenizer.nval);
} else return tokenizer.sval;
} catch (Exception e) {
e.printStackTrace();
return null;
}
}

private boolean input()
{
s = next();
return true;
}

private boolean isPalindrom(int start, int end)
{
while (start <= end) {
if (s.charAt(start) != s.charAt(end)) return false;
start++;
end--;
}

return true;
}

private void solve()
{
int len = s.length(), i, j;
dp = new int[len + 1];
dp[0] = 0;
for (i = 1; i <= len; i++) {
for (dp[i] = N, j = 1; j <= i; j++) {
if (isPalindrom(j - 1, i - 1)) {
dp[i] = Math.min(dp[i], dp[j - 1] + 1);
}
}
}

cout.println(dp[len]);
cout.flush();
}

public void run()
{
init();

int n = Integer.parseInt(next());
while (n-- > 0) {
input();
solve();
}
}

public static void main(String[] args)
{
new Thread(new Main()).start();
}
}