poj1159 Palindrome

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本文详细解析了 POJ1159 Palindrome 的动态规划解决方案，介绍了如何通过动态规划算法确定将任意字符串转换为回文串所需的最小插入次数，并提供了两种实现方式的代码示例。

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poj1159 Palindrome

Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 68871 Accepted: 23964
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

比较经典的一道动态规划题，状态转移方程容易寻找：
dp[i][j]=min(dp[(i+1)][j],dp[i][j-1])+1;
可以这样理解，dp[i][j]表示第i个字符到第j个字符边为回文串所需最少操作步数
当s[i] == s[j] 很明显dp[i][j] = dp[i+1][j-1]
当s[i]!=s[j] 可以在i处插入s[j],也可以在j处插入s[i],则有状态转移方程
dp[i][j]=min{dp[i+1][j]+1,dp[i][j-1]+1};
附上AC代码

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

#define MAXN 5005

short dp[MAXN][MAXN];
char str[MAXN];

int main(){

    int N;
    int i,j;

    while(~scanf("%d",&N)){
        scanf("%s",str);

        memset(dp,0,sizeof(dp));

        for(i=N-1;i>=0;--i){
            dp[i][i]=0;
            for(j=i+1;j<N;++j){
                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]);
    }

    return 0;
}

当然，很明显，根据状态转移方程也可以用滚动数组进行空间复杂度优化

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

#define MAXN 5005

short dp[2][MAXN];//滚动数组
char str[MAXN];

int main(){

    int N;
    int i,j;

    while(~scanf("%d",&N)){
        scanf("%s",str);

        memset(dp,0,sizeof(dp));

        for(i=N-1;i>=0;--i){
            dp[i%2][i%2]=0;
            for(j=i+1;j<N;++j){
                if(str[i]==str[j]){
                    dp[i%2][j]=dp[(i+1)%2][j-1];
                }
                else{
                    dp[i%2][j]=min(dp[(i+1)%2][j],dp[i%2][j-1])+1;
                }
            }
        }

        printf("%d\n",dp[0][N-1]);
    }

    return 0;
}