Palindrome
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 59029 | Accepted: 20505 |
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
题意是给你一个字符串，让你最少添加多少个字符使得字符串为一个回文串，参考了网上大神们的想法 首先把原始的串反转，然后求这两个串的最长公共子序列，
需要添加的字符=字符串的长度-公共字符串
感觉很好理解，然后写一下最长公共子序列就好了，简单的dp啊
还有中间需要注意的地方，由于题目要求的内存为65536K，要是使用int保存
dp[5005][5005]，毫无疑问内存超限了，好多大神写的是滚动数组。本弱弱利用short来存储dp数组，险过啊！！！由于int保存为4个字节，而short只用2个字节来保存，你懂得。
//神奇的short，不会超内存，
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
short dp[5005][5005];
char s1[5005],s2[5005];
int main(){
int n,i,j;
int ans;
while(~scanf("%d",&n)){
scanf("%s",s1);
int l = strlen(s1);
for(i=0;i<l;i++){
s2[l-i-1]=s1[i];
}
s2[l] = '\0';
memset(dp,0,sizeof(dp));
for(i=1;i<=l;i++){
for(j=1;j<=l;j++){
if(s1[i-1]==s2[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",l-dp[l][l]);
}
return 0;
}