1033 - Generating Palindromes（傻DP）

最小成本回文生成器

最新推荐文章于 2020-07-20 15:05:46 发布

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本文介绍了一种算法，该算法能够确定将任意给定字符串转换为回文所需的最少字符插入数量。通过动态规划方法，文章详细解释了如何计算这一最小成本，并提供了完整的C++实现代码。

By definition palindrome is a string which is not changed when reversed. “MADAM” is a nice example of palindrome. It is an easy job to test whether a given string is a palindrome or not. But it may not be so easy to generate a palindrome.

Here we will make a palindrome generator which will take an input string and return a palindrome. You can easily verify that for a string of length n, no more than (n - 1) characters are required to make it a palindrome. Consider “abcd” and its palindrome “abcdcba” or “abc” and its palindrome “abcba”. But life is not so easy for programmers!! We always want optimal cost. And you have to find the minimum number of characters required to make a given string to a palindrome if you are only allowed to insert characters at any position of the string.

Input
Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a string of lowercase letters denoting the string for which we want to generate a palindrome. You may safely assume that the length of the string will be positive and no more than 100.

Output
For each case, print the case number and the minimum number of characters required to make string to a palindrome.

Sample Input
Output for Sample Input
6
abcd
aaaa
abc
aab
abababaabababa
pqrsabcdpqrs
Case 1: 3
Case 2: 0
Case 3: 2
Case 4: 1
Case 5: 0
Case 6: 9

最少填多少个字符成回文。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int t;
char s[105];
int dp[105][105];
int dfs(int l,int r)
{
    if(dp[l][r] != -1)return dp[l][r];
    if(l >= r)return dp[l][r] = 0;
    if(s[l] == s[r])
        return dp[l][r] = dfs(l + 1,r - 1);
    else
        return dp[l][r] = min(dfs(l + 1,r),dfs(l,r - 1)) + 1;
}
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    int kase = 1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        int len = strlen(s);
        memset(dp,-1,sizeof(dp));
        printf("Case %d: %d\n",kase++,dfs(0,len - 1));
    }
    return 0;
}