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Palindrome
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 52910 | Accepted: 18248 |
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
说白了就是求和自己相反字符串的lcs
short可以卡过
还是老老实实写滚动数组吧
#include<cstdio>
#include<cstring>
int dp[2][5050];
char str1[5050],str2[5050];
int maxx(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n,i,j;
int l;
while(scanf("%d",&n)!=EOF)
{
scanf("%s",str1+1);
for(i=1;i<=n;i++)
{
str2[n+1-i]=str1[i];
}
memset(dp,0,sizeof(dp));
int flag=1;
int xx=0;
for(i=1;i<=n;i++)
{
flag=1-flag;
for(j=1;j<=n;j++)
{
if(str1[i]==str2[j])
{
dp[flag][j]=dp[1-flag][j-1]+1;
}
else
{
dp[flag][j]=maxx(dp[flag][j-1],dp[1-flag][j]);
}
xx=maxx(xx,dp[flag][j]);
// printf("%d ",dp[flag][j]);
}
// puts("");
}
printf("%d\n",n-xx);
}
}
poj 1159 Palindrome lcs+滚动数组