杭电 1513 Palindrome LCS DP

最新推荐文章于 2018-03-01 15:05:18 发布

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本文介绍了一种使用动态规划中的滚动数组方法解决构造最短回文串的问题。通过一个具体示例，详细展示了如何计算最少需要插入多少字符使得任意字符串变成回文串。

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Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3955 Accepted Submission(s): 1349

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

第一次用滚动数组，感觉自己萌萌哒QAQ。

嘛，滚动数组其实并不难，关键是要用i来确定，这样会比较方便点。虽然还是有点小不理解吧，不过这都不是事儿~

#include<stdio.h>
#include<string.h>
#define maxn 5000+10
#define max(a,b) (a>b?a:b)
char s1[maxn],s2[maxn];
int dp[2][maxn];

int main(){
	int n;
	while(~scanf("%d",&n)){
		scanf("%s",s1);
		strcpy(s2,s1);
		strrev(s2);
		memset(dp,0,sizeof(dp));
		int i,j,x=0,y=1;
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				x=i%2;y=1-x;
				if(s1[i-1]==s2[j-1])
				dp[x][j]=dp[y][j-1]+1;
				else
				dp[x][j]=max(dp[y][j],dp[x][j-1]);
			}
		}
		printf("%d\n",n-dp[n%2][n]);
	}
	return 0;
}