poj1159 Palindrome

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 60770		Accepted: 21174

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

2

Source

IOI 2000

题目链接:http://poj.org/problem?id=1159

题目意思:给出一个字符串(n为其长度),问插入多少个字符就能使其变为回文串.

思路一:设dp[i][j]表示字符c[i] - c[j]变为回文的最小需要的字符个数,那么就有if(c[i] == c[j]) dp[i][j] = dp[i-1][j+1]  else dp[i][j] = min(dp[i+1][j], dp[i][j-1])+1

提示:dp要开short类型不然会TLE.

代码如下:

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

const int inf = 0x3f3f3f3f;

int n;
short int dp[5005][5005];
char c[5005];

int dfs(int l, int r){
    if(l >= r) return 0;
    if(~dp[l][r]) return dp[l][r];
    short int& res = dp[l][r];
    res = inf;
    if(c[l] == c[r]) res = dfs(l+1, r-1);
    else
        res = min(dfs(l+1, r), dfs(l, r-1)) + 1;
    return res;
}

int main(){
    scanf("%d", &n);
    scanf("%s", c);
    memset(dp, -1, sizeof(dp));
    printf("%d\n", dfs(0, n-1));
    return 0;
}

还有一种思路:设原序列S的逆序列为S' , 最少需要补充的字母数 = 原序列S的长度 —  S和S'的最长公共子串长度(这个是看别人的学到的),

然后就是TLE的解决方法了   开个二位的滚动数组就行了。

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

int n, dp[2][5005];
char c[5005];

int main(){
    scanf("%d", &n);
    scanf("%s", c+1);
    memset(dp, 0, sizeof(dp));
    int tmp = 0;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            if(c[i] == c[n-j+1]) dp[i%2][j] = dp[(i-1)%2][j-1] + 1;
            else dp[i%2][j] = max(dp[i%2][j-1], dp[(i-1)%2][j]);
            tmp = max(tmp, dp[i%2][j]);
        }
    }
    printf("%d\n", n-tmp);
    return 0;
}