算法设计与复杂度 模拟考 Palindrome

总时间限制: 

3000ms 

内存限制: 

65536kB

描述

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.

输入

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.

输出

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

样例输入

5
Ab3bd

样例输出

2

来源

IOI 2000

可以采用直接dp，dp[i][j] = min {以下三种选择}dp[i+1][j-1] （如果str[i] == str[j]），dp[i+1][j]+1 （右边强行添加一个）， dp[i][j-1]+1 （左边强行添加一个）

也可以转换成公共最长子序列问题。

第一种方法代码

注意取余数的时候不要用减法，余数存在为负数的情况。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <string.h>

using namespace std;

int n;
char letter[5005];
int num[5005][3];

int main()
{
    scanf("%d", &n);
    scanf("%s", letter);
    for(int j = 0; j < n; j++)
    {
        num[j][0] = 0;
        if(j + 1 < n && letter[j] != letter[j + 1])
            num[j][1] = 1;
        else
            num[j][1] = 0;
    }
    int roll = 2;
    for(int i = 2; i < n; i++)
    {
        for(int j = 0; i + j < n; j++)
        {
            if(letter[j] == letter[j + i])
                num[j][roll] = num[j + 1][(roll + 1) % 3];
            else
                num[j][roll] = min(num[j + 1][(roll + 2) % 3], num[j][(roll + 2) % 3]) + 1;
            //printf("%c %c\n", letter[j], letter[j + i]);
            //printf("%d %d %d %d\n", j, i, roll, num[j][roll]);
        }
        roll = (roll + 1) % 3;
    }
    printf("%d\n",num[0][(roll + 2) % 3]);
    return 0;
}

最长公共子序列

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <string.h>

using namespace std;

int n;
char str1[5005], str2[5005];
int dp[5005][2];

int main()
{
    scanf("%d", &n);
    scanf("%s", str1);
    for(int i = 0; i < n; i++)
    {
        str2[i] = str1[n - 1 - i];
    }
    memset(dp, 0, sizeof(dp));
    int roll = 1;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            if(str1[i - 1] == str2[j - 1])
                dp[j][roll] = dp[j - 1][roll ^ 1] + 1;
            else
                dp[j][roll] = max(dp[j][roll ^ 1], dp[j - 1][roll]);
        }
        roll = roll ^ 1;
    }
    printf("%d\n", n - dp[n][roll ^ 1]);
    return 0;
}