poj1159（回文字串）

Palindrome

Time Limit: 3000MSMemory Limit: 65536KTotal Submissions: 41514Accepted: 14146

Description

A palindrome is a symmetricalstring, that is, a string read identically from left to right aswell as from right to left. You are to write a program which, givena string, determines the minimal number of characters to beinserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can betransformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However,inserting fewer than 2 characters does not produce apalindrome.

Input

Your program is to read fromstandard input. The first line contains one integer: the length ofthe input string N, 3 <= N <= 5000.The second line contains one string with length N. The string isformed from uppercase letters from 'A' to 'Z', lowercase lettersfrom 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercaseletters are to be considered distinct.

Output

Your program is to write tostandard output. The first line contains one integer, which is thedesired minimal number.

Sample Input

5Ab3bd

Sample Output

2

Source

IOI 2000

 

#include<stdio.h>
#include<string.h>
short int d[5010][5010];//如果不用short,会超内存
char s[5010];
int min(int x,int y)
{
 return x<y?x:y;
}
int main()
{
 int n,i,j;
 scanf("%d",&n);
 scanf("%s",s);
 memset(d,0,sizeof(d));
 for(i=n-1;i>=0;i--)
  for(j=i+1;j<n;j++)//如果两个游标所指字符相同，向中间缩小范围
  {
   if(s[i]==s[j])
    d[i][j]=d[i+1][j-1];
   else//如果不同，典型的状态转换方程
    d[i][j]=min(d[i+1][j],d[i][j-1])+1;
  }
  printf("%d\n",d[0][n-1]);
 return 0;
}

 

这题还可以用滚动数组做（很多大牛们都是这样做的）

参考资料http://archive.cnblogs.com/a/2155738/

下面是我照这别人代码敲的。。输入格式很容易错哦。。WA了好多次了。。就是scanf()这边。。还是建议用cin>>s1吧。。。下面代码是WA的。。把输入格式改一下就A了。。

 #include<stdio.h>
#include<string.h>
#define max 5005
char s1[max],s2[max];
int maxlen[2][max];
int main()
{
 int n,i,j;
 while(scanf("%d",&n)!=EOF)
 {
  //memset(s1,0,sizeof(s1));
  //memset(s2,0,sizeof(s2));
  
  for(i=1;i<=n;i++)
   scanf("%c",&s1[i]);
  s1[n+1]='\0';
  for(i=1;i<=n;i++)
   s2[i]=s1[n-i+1];
  memset(maxlen,0,sizeof(maxlen));
  int e=0;
  for(i=1;i<=n;i++)
  {
   e=1-e;
   for(j=0;j<=n;j++)
   {
    if(s1[i]==s2[j])
     maxlen[e][j]=maxlen[1-e][j-1]+1;
    else
    {
     intlen1=maxlen[e][j-1];
     intlen2=maxlen[1-e][j];
     if(len1>len2)maxlen[e][j]=len1;
     else
      maxlen[e][j]=len2;
    }
   }
  }
  printf("%d\n",n-maxlen[e][n]);
 }
 return 0;
}