添字成回文词 POJ1159

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添字成回文词 POJ1159

2013年01月15日 ⁄ 综合 ⁄ 共 2838字 ⁄ 字号小中大 ⁄ 评论关闭

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 39691		Accepted: 13485

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

题意：

求出将给定字符串变成回文词所需要插入的最少字符数，并输出任意给定的字符串的回文词。

分析：

首先拿一个字符串AbDCb观察，最优解为AbCDCbA。可以发现其对应关系：A b - D C b - 与 A b C D C b A。采用LCS（Longest Common Subsequence，最长公共子串-可不连续）算法求出最大重叠的部分。利用一个LenMatrix[N][N]二维数组记录下重叠长度，利用一个DirectFlag[N][N]记录下长度增长路线，利用record[LEN]记录下取得最长子串所覆盖的字符。输出完整的回文数要利用record数组从两边向中间靠拢，直到全部输出。

POJ1159只要求求出所需最少的字符数，下面的代码同时将最后生成的回文词也输出了。

程序源代码如下：

#include <iostream> 
using namespace std; 

#define LEN 100 

char array[LEN]={' '},inver[LEN]={' '}; 
char HuiWen[LEN*2];
bool record[LEN]={0};
short LenMatrix[LEN][LEN] = {0};
char DirectFlag[LEN][LEN] = {0}; 

short getLCSLength(int n) 
{ 
	int i,j;
    for (i=1;i<=n;++i) 
    { 
        for (j=1;j<=n;++j) 
        { 
            if (array[i]==inver[j]) 
            { 
                LenMatrix[i][j] = LenMatrix[i-1][j-1] + 1; 
                DirectFlag[i][j] = 1; 
            } 
            else if (LenMatrix[i-1][j] >= LenMatrix[i][j-1]) 
            { 
                LenMatrix[i][j] = LenMatrix[i-1][j]; 
                DirectFlag[i][j] = 2; 
            } 
            else 
            { 
                LenMatrix[i][j] = LenMatrix[i][j-1]; 
                DirectFlag[i][j] = 3; 
            } 
        }
    }
	return LenMatrix[n][n];
} 

void BackRecord(int i, int j) 
{ 
    if(!i || !j) return; 
    switch(DirectFlag[i][j]) 
    { 
    case 1: 
        BackRecord (i-1,j-1); 
     	record[i]=true;
        break; 
    case 2: 
        BackRecord (i-1,j); 
	    break; 
    case 3: 
        BackRecord (i,j-1); 
	} 
} 

void showWhole(int n,int min)
{
	int i=1;
	int j=n;
	int k=0;
	int half=min/2;
	while(1)
	{
		while(record[i])
		{
			HuiWen[k]=array[i];
			HuiWen[min-1-k]=array[i];
			k++;
			i++;
			j--;
		}
		if(k>half)break;
		while(!record[i])
		{
			HuiWen[k]=array[i];
			HuiWen[min-1-k]=array[i];
			k++;
			i++;
		}
		if(k>half)break;
		while(record[j])
		{
			HuiWen[k]=array[j];
			HuiWen[min-1-k]=array[j];
			k++;
			i++;
			j--;
		}
		if(k>half)break;
		while(!record[j])
		{
			HuiWen[k]=array[j];
			HuiWen[min-1-k]=array[j];
			k++;
			j--;
		}
		if(k>half)break;
	}
	HuiWen[min]='\0';
	cout<<HuiWen<<endl;
}
int main() 
{ 
	freopen("in.txt","r",stdin);
	int n;
	cin>>n;
	cin>>array+1;
	int i;
	for(i=1;i<=n;i++)inver[i]=array[n+1-i];  //字符串倒置
	inver[i]='\0';

    short comLen=getLCSLength (n);  //利用矩阵累加法得到最长子串
      
	if(comLen==1)	//没有公共子串
	{		
		cout<<n-1<<endl;
		cout<<array+1<<inver+2<<endl;
		return 0;
	}else 
	{  
		BackRecord(n,n);   //后退，记录下最长公共子串的路线
		cout<<n-comLen<<endl;
		showWhole(n,n+n-comLen);
	}
    return 0; 
}

测试用例：

dA432b3b11234d

输出结果：