10453 Make Palindrome （dp)

最新推荐文章于 2021-06-09 15:48:00 发布

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最新推荐文章于 2021-06-09 15:48:00 发布

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原文链接：http://www.cnblogs.com/james1207/p/3327364.html

本文介绍了一种算法，用于解决给定字符串通过插入最少字符生成回文串的问题。通过对字符串进行动态规划，确定最少插入字符数并生成相应的回文串。文章提供了详细的代码实现及解析。

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Problem A

Make Palindrome

Input: standard input

Output: standard output

Time Limit: 8 seconds

By definition palindrome is a string which is not changed when reversed. "MADAM" is a nice example of palindrome. It is an easy job to test whether a given string is a palindrome or not. But it may not be so easy to generate a palindrome.

Here we will make a palindrome generator which will take an input string and return a palindrome. You can easily verify that for a string of length 'n', no more than (n-1) characters are required to make it a palindrome. Consider "abcd" and its palindrome "abcdcba" or "abc" and its palindrome "abcba". But life is not so easy for programmers!! We always want optimal cost. And you have to find the minimum number of characters required to make a given string to a palindrome if you are allowed to insert characters at any position of the string.

Input

Each input line consists only of lower case letters. The size of input string will be at most 1000. Input is terminated by EOF.

Output

For each input print the minimum number of characters and such a palindrome seperated by one space in a line. There may be many such palindromes. Any one will be accepted.

Sample Input

abcdaaaaabcaababababaabababapqrsabcdpqrs

Sample Output

3 abcdcba0 aaaa2 abcba1 baab0 abababaabababa9 pqrsabcdpqrqpdcbasrqp

题意：给定字符串。可以在任意位置增添字符，求最少步骤生成回文串。以及生成的串

思路：

和这题一样。http://blog.youkuaiyun.com/accelerator_/article/details/11542037。

多开一个vis数组记录状态转移方式便于输出。

代码：

#include <stdio.h>
#include <string.h>
const int MAXN = 1005;

char sb[MAXN];
int dp[MAXN][MAXN], vis[MAXN][MAXN], n, i, j;

void print(int i, int j) {
	if (i == j) {
		printf("%c", sb[i]);
		return;
	}
	if (i > j)
		return;
	if (vis[i][j] == -1) {
		printf("%c", sb[i]);
		print(i + 1, j - 1);
		printf("%c", sb[j]);
	}
	else if (vis[i][j] == 0) {
		printf("%c", sb[i]);
		print(i + 1, j);
		printf("%c", sb[i]);
	}
	else if (vis[i][j] == 1) {
		printf("%c", sb[j]);
		print(i, j - 1);
		printf("%c", sb[j]);
	}
}

int main() {
	while (gets(sb) != NULL) {
		n = strlen(sb);
		for (i = n - 1; i >= 0; i --) {
			for (j = i + 1; j < n; j ++) {
				if (sb[i] == sb[j]) {
					dp[i][j] = dp[i + 1][j - 1];
					vis[i][j] = -1;
				}
				else {
					if (dp[i + 1][j] < dp[i][j - 1]) {
						dp[i][j] = dp[i + 1][j] + 1;
						vis[i][j] = 0;
					}
					else {
						dp[i][j] = dp[i][j - 1] + 1;
						vis[i][j] = 1;
					}
				}
			}
		}
		printf("%d ", dp[0][n - 1]);
		print(0, n - 1);
		printf("\n");
	}
	return 0;
}

转载于:https://www.cnblogs.com/james1207/p/3327364.html