11584 - Partitioning by Palindromes(dp)

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

#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
#define min(a,b) ((a)<(b)?(a):(b))
const int N = 1005;

int t, n, dp[N], ok[N][N];
char str[N];

void init() {//预处理
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= i; j++) {
			if (i == j || (str[i] == str[j] && (ok[j + 1][i - 1] || i - j == 1)))
				ok[j][i] = 1;
			else ok[j][i] = 0;
		}
}

int solve() {
	memset(dp, INF, sizeof(dp));
	dp[0] = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= i; j++) {
			if (ok[j][i]) {
				dp[i] = min(dp[i], dp[j - 1] + 1);
			}
		}
	return dp[n];
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%s", str + 1);
		n = strlen(str + 1);
		init();
		printf("%d\n", solve());
	}
	return 0;
}

11584 - Partitioning by Palindromes(dp)

原文：http://blog.csdn.net/accelerator_/article/details/19292231