Palindrome
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 52966 | Accepted: 18271 |
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
| 13368265 | happystick | 1159 | Accepted | 148K | 719MS | C++ | 720B | 2014-08-23 21:42:27 |
//最少插入字符个数 =字符串长度-正、逆序的最长公共子序列长度
/*
一维数组+short 优化wa了好几次，无语了 刚开始是以为多组测试数据，一直wa看了半天人家的代码才知道。。。。就一组测试数据不过这个代码应该是最优代码吧 参考的南阳优秀代码 同样代码在hdoj提交wa 无语
Time:2014-8-23 21:47
*/
#include<stdio.h>
#include<string.h>
int main(){
char s1[5005],s2[5005];
short dp[5005];
int N;
scanf("%d",&N);
scanf("%s",s1);
for(int i=0;i<N;i++){
s2[i]=s1[N-i-1];
};s2[N]='\0';
// printf("%s %s\n",s1,s2);
int old,temp;
memset(dp,0,sizeof(dp));
//old 代表dp[i-1][j-1]
//dp[j]代表 dp[i-1][j]
for(int i=0;i<N;i++){
old=0;//将old初始置为 0
for(int j=0;j<N;j++){
temp=dp[j];//将dp[i-1][j]记下来，为下一列的old
if(s1[i]==s2[j]){
dp[j]=old+1;
}else{
dp[j]=temp>dp[j-1]?temp:dp[j-1];//取dp[i-1][j]和dp[i][j-1]中的较大值
}
old=temp;//上一列的dp[i-1][j]为下一列的dp[i-1][j-1]
}
}
printf("%d\n",N-dp[N-1]);
return 0;
}