HDU 1513 最长子序列

Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3641 Accepted Submission(s): 1252

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

Ab3bd

Sample Output

Source

IOI 2000

题目意思：

给一个长度为n的字符串，问最少添加多少字符使得该字符串成为回文串。

思路：

把原串倒过来，两个串取最长子序列m，然后n-m即为加的字符数。子序列长度最长，意味着添加的字符数最少。

由于n太大，需要用滚动数组。

dp[i&1][j]表示dp[i][j]; dp[!(i&1)][j]表示dp[i-1][j]

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <vector>
 6 #include <queue>
 7 #include <cmath>
 8 #include <set>
 9 using namespace std;
10 
11 #define N 5005
12 
13 int max(int x,int y){return x>y?x:y;}
14 int min(int x,int y){return x<y?x:y;}
15 int abs(int x,int y){return x<0?-x:x;}
16 
17 
18 int dp[2][N];
19 int n;
20 char s[N];
21 
22 main()
23 {
24     int i, j, k;
25     while(scanf("%d",&n)==1){
26         memset(dp,0,sizeof(dp));
27         scanf("%s",s+1);
28         char s1[N];
29         for(i=n;i>=1;i--) s1[i]=s[n-i+1];
30         int maxh=0;
31         for(i=1;i<=n;i++){
32             memset(dp[i&1],0,sizeof(dp[i&1]));
33             for(j=1;j<=n;j++){
34                 if(s[i]==s1[j]){
35                     dp[i&1][j]=max(dp[!(i&1)][j-1]+1,dp[i&1][j]);
36                 }
37                 else dp[i&1][j]=max(dp[!(i&1)][j],dp[i&1][j-1]);
38                 maxh=max(maxh,dp[i&1][j]);
39             }
40         }
41         printf("%d
",n-maxh);
42     }
43 }