Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4751    Accepted Submission(s): 1625

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

5
Ab3bd

Sample Output

2

Source

IOI 2000

这题用区间DP会爆内存。。正确解法利用LCS解,并且LCS算法算的时候只和当前行和上一行有关,所以利用滚动数组压缩空间

///题意:将一个字符串变成回文串需要添加的最少字符数量
///解法:将字符串反过来.求出原字符串与现在的字符串的LCS，然后用原串减掉LCS的长度即为最少添加的字符
///这个解法的好处就是可以将DP数组变成滚动数组
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = 5001;
int dp[2][N];
char str[N],str1[N];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        scanf("%s",str+1);
        for(int i=1;i<=n;i++){
            str1[n-i+1]=str[i];
        }
        ///printf("%s",str1+1);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(str[i]==str1[j]){
                    dp[i%2][j] = dp[(i-1)%2][j-1]+1;
                }else{
                    dp[i%2][j] = max(dp[i%2][j-1],dp[(i-1)%2][j]);
                }
            }
        }
        printf("%d\n",n-max(dp[1][n],dp[0][n]));
    }
}

贴个爆内存的代码。。开short都没用,hdu数据强吗 ORZ ~~~

///题意:将一个字符串变成回文串需要添加的最少字符数量
///解法:区间DP，dp[i][j]代表区间i-j里面要添加多少个字符才能将其变成回文串
///如果 str[i] == str[j] 那么 dp[i][j] = dp[i+1][j-1]
///否则 dp[i][j] = min(dp[i+1][j] ,dp[i][j-1])+1 即考虑是在第i+1个字符之前添加str[j]还是在
///在j-1之后添加字符str[i]
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = 5000;
short dp[N][N];
char str[N];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        scanf("%s",str);
        for(int i=1;i<=n;i++) dp[i][i]=0;///区间长度为1的时候不需要添加就是回文串
        for(int l = 2;l<=n;l++){
            for(int i=0;i<=n-l+1;i++){
                int j = i+l-1;
                if(str[i]==str[j]) {
                    dp[i][j] = dp[i+1][j-1];
                }else{
                    dp[i][j] = min(dp[i+1][j],dp[i][j-1])+1;
                }
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
}