回文串匹配——POJ 1159

对应POJ 题目：点击打开链接

Palindrome

Description

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted
into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters
from ‘A‘ to ‘Z‘, lowercase letters from ‘a‘ to ‘z‘ and digits from ‘0‘ to ‘9‘. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5
Ab3bd

Sample Output

2

题意：给出长度为n的字符串，求最少添加多少个字符就能把原字符串变成回文串（添加的位置是任意的）

思路：最少添加数量 = 原串长度 - 原串跟其逆序串的最长公共子序列长度。

证明：原串跟其逆序串的最长公共子序列 = 原串的最长不连续回文串（比如 aABdCBcAm 的最长不连续回文串是ABCBA），既然最长不连续回文串知道了，那在剩下的每个字符的对称位置添加该字符，就能把原串变成回文串。而剩下的字符个数也就是题目的答案，接下来不言而喻。。。

要注意的是，求最长公共子序列时需要把空间压缩成一维。使用到一些技巧：

0    1    2    3    4    5   j ->

0   0    0    0    0    ...

1   0    a    b    ...

2   0    c    d    ...

i    0    ...

dp[i][j] 表示A串前i个字符跟B串前j个字符的最长公共子序列。转移方程为

if(A[i-1] == B[j-1]) 
dp[i][j] = dp[i-1][j-1]  + 1;

else dp[i][j] = max(dp[i][j-1], dp[i-1][j]);

通俗地讲就是如果A[i] == B[j]，那dp[i][j] 就等于它左上角的值加1，否则dp[i][j] 就等于它左边跟它上面的值中较大的那个。

仔细想想我们就能把不必要的空间消耗省掉，具体看代码。。。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#define M 5010
using namespace std;
char s1[M];
char s2[M];
int dp[M];

int main()
{
	//freopen("in.txt", "r", stdin);
	int n;
	int i, j, t, tmp;
	while(~scanf("%d", &n))
	{
		memset(dp, 0, sizeof(dp));
		scanf("%s", s1);
		for(i=0; i<n; i++)
			s2[n-i-1] = s1[i];
		for(i=0; i<n; i++){
            tmp=0;
            for(j=0; j<n; j++){
                t=dp[j+1];
                if(s1[i] == s2[j]) dp[j+1] = tmp+1;
                else if(dp[j] > t) dp[j+1] = dp[j];
                tmp=t;
            }
        }
		printf("%d\n", n - dp[n]);
	}
	return 0;
}