HDOJ-----1513Palindrome(LCS)

最新推荐文章于 2017-09-12 16:02:25 发布

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本文介绍了一种算法，用于确定将任意字符串转换为回文串所需的最少字符插入数量。通过计算原始字符串与其逆序字符串之间的最长公共子序列长度，进而得出所需插入的字符数。

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Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5244 Accepted Submission(s): 1798

Problem 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

球最少再插入几个字符可以构成回文串，显然最多加n个，即与原串逆序的一个串加到末尾必定回文

也就是求原串与逆序串最长公共子序列，即不需要变动的长度，再用原串长度减去就是需要插入的

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 10010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
char st[maxn], str[maxn];
int dp[2][maxn];
int main(){
	int ans, n, l1, l2;
	while(~scanf("%d", &n)){
        scanf("%s", st);
        for(int i = 0; i < n; i++){
            str[n-1-i] = st[i];
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(st[i-1] == str[j-1]){
                    dp[i&1][j] = dp[(i-1)&1][j-1]+1;
                }
                else{
                    dp[i&1][j] = max(dp[(i-1)&1][j], dp[i&1][j-1]);
                }
            }
        }
        printf("%d\n", n-dp[n&1][n]);
	}
	return 0;
}