POJ1159求LCS长度 滚动数组优化空间

                                                                      Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 64652 Accepted: 22566

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

5Ab3bd

Sample Output

2

题目意思

插入最少的字符到一个字符串中使得这个字符串是“回文字符串”。

题解

答案为：原字符串的长度 — （原字符串和逆字符串的最长公共子序列的长度）

由于数据是5000*5000的，所以开dp[n][n]的数组不行，利用滚动数组只需要dp[2][n]

if(s1[i] == s2[j])  dp[(i+1)%2][j+1] = dp[i%2][j]+1;            else dp[(i+1)%2][j+1] = max(dp[(i+1)%2][j], dp[i%2][j+1]);答案为：

int ans = n - dp[n%2][n];

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int  maxn = 5010;int dp[2][maxn];char s1[maxn], s2[maxn];int n;int main(){    scanf("%d", &n);    scanf("%s", s1);    for(int i = 0; i < n; i++) {        s2[n-i-1] = s1[i];    }    for(int i = 0; i < n; i++) {        for(int j = 0; j < n; j++) {            if(s1[i] == s2[j])  dp[(i+1)%2][j+1] = dp[i%2][j]+1;            else dp[(i+1)%2][j+1] = max(dp[(i+1)%2][j], dp[i%2][j+1]);        }    }    int ans = n - dp[n%2][n];    printf("%d\n", ans);    return 0;}