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POJ 1159-Palindrome（DP/LCS变形）

2024-08-02 11:19:52   218人阅读

Palindrome
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 53770 Accepted: 18570

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：求一个字符串变成回文串至少要添加多少个字符（可以在任意位置添加） 
思路：将串s逆转得到s’ （注意strrev()会CE sad） ，然后求s和s‘的最长公共子序列（LCS）  n-LCS即为答案。
很吃惊? 其实是这样的：因为s和s’的最长公共子序列肯定是回文，所以剩下的长度只需要添加n-LCS个字符既可以满足总体回文了。。想象成添加可能不好理解，可以这么想，LCS部分的字符串已经是回文了，所以我们不需要管它了，然后就是把剩余的字符插空了，举个例子,比如 Ab3bd 反转后得 s’ db3bA 所以LCS=3 （b3b） 那么剩余的两个字符分别为 A和d 现在我们依次把他们插入原串s：因为A在最左边，所以在最右边插入一个A 得Ab3bdA 然后此时的在右边数第二位，所以我们在左边数第二位插入一个d 得 Adb3bdA  ok大功告成。。。
 至于求LCS。。蒟蒻只会o（n*n） 然后数组要开short才能过。。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define maxn 360
#define pp pair<int,int>
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n;
short dp[5005][5005]={0};
char s[5010],t[5010];
void solve()
{
	for(int i=0;i<n;i++)
		t[i]=s[n-i-1];
	t[n]='\0';
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		if(s[i-1]==t[j-1])
		dp[i][j]=dp[i-1][j-1]+1;
		else
		dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
	printf("%d\n",n-dp[n][n]);
}
int main()
{
	scanf("%d",&n);
	scanf("%s",s);
	solve();
	return 0;
}

POJ 1159-Palindrome（DP/LCS变形）


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