标签：palindrome hdoj 1513 lcs dp
Palindrome
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4194 Accepted Submission(s): 1427
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
Sample Output
倒转字符串，然后与原字符串求最长公共子序列，然后用长度减就行了；
但这个题目要求字符串为5000，所以二维数组肯定爆内存，所以用滚动数组，用dp[2][5500]即可！
#include<stdio.h>
#include<string.h>
#define N 5500
int dp[2][N];
char s1[N],s2[N];
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(dp,0,sizeof(dp));
int len,i,j;
getchar();
scanf("%s",s1);
len=strlen(s1);
for(i=len-1,j=0;i>=0;i--)//字符串倒序
{
s2[i]=s1[j++];
}
for(i=1;i<=len;i++)
for(j=1;j<=len;j++)
{
int x=i%2;//在两行之间 循环进行
int y=1-x;
if(s1[i-1]==s2[j-1])
dp[x][j]=dp[y][j-1]+1;
else
dp[x][j]=max(dp[x][j-1],dp[y][j]);
}
printf("%d\n",len-dp[len%2][len]);
}
return 0;
}
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hdoj 1513 Palindrome
标签：palindrome hdoj 1513 lcs dp
原文地址：http://blog.csdn.net/zhangxiaoxiang123/article/details/47831895
