CodeForces 666A Reberland Linguistics(DP)

本文探讨了Reberland语言中单词构造规则及其应用,特别关注于如何识别长度为2或3的可能后缀。通过实例解析,详细介绍了后缀的构造过程和识别方法。

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A. Reberland Linguistics
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.

For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word — some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction — it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).

Here is one exercise that you have found in your task list. You are given the word s. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.

Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.

Let's look at the example: the word abacabaca is given. This word can be obtained in the following ways: , where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {aca, ba, ca}.

Input

The only line contains a string s (5 ≤ |s| ≤ 104) consisting of lowercase English letters.

Output

On the first line print integer k — a number of distinct possible suffixes. On the next k lines print suffixes.

Print suffixes in lexicographical (alphabetical) order.

Examples
input
abacabaca
output
3
aca
ba
ca
input
abaca
output

0


动态规划,一开始看错了题目,题目的意思是不能有连续相邻的两个后缀是相同的


#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <stdio.h>
#include <map>
#include <string>

using namespace std;
#define MAX 10000
struct Node
{
  string a;
}a[MAX*10];
int cmp(Node a,Node b)
{
    return a.a<b.a;
}
int dp[MAX+5];
string s;
map<string,int> m;
int main()
{
    cin>>s;
    int len=s.length();
    memset(dp,0,sizeof(dp));
    dp[len]=1;dp[len-1]=0;dp[len+1]=0;
    m.clear();
    int cnt=0;
    for(int i=len-2;i>=5;i--)
    {
        if(dp[i+2]!=0)
        {
           string  b=s.substr(i,2);
            if(dp[i+2]!=2||b!=s.substr(i+2,2))
            {
                if(!m.count(b))
                {
                     m[b]=1;
                     a[cnt++].a=b;
                }
                dp[i]+=2;
            }
        }
         if(dp[i+3]!=0)
        {
            string b=s.substr(i,3);
            if(dp[i+3]!=3||b!=s.substr(i+3,3))
            {
                if(!m.count(b))
                {
                     m[b]=1;
                     a[cnt++].a=b;
                }
                dp[i]+=3;
            }
        }
    }
    sort(a,a+cnt,cmp);
    printf("%d\n",cnt);
    for(int i=0;i<cnt;i++)
        cout<<a[i].a<<endl;
    return 0;

}


区间DP是一种动态规划的方法,用于解决区间范围内的问题。在Codeforces竞赛中,区间DP经常被用于解决一些复杂的字符串或序列相关的问题。 在区间DP中,dp[i][j]表示第一个序列前i个元素和第二个序列前j个元素的最优解。具体的转移方程会根据具体的问题而变化,但是通常会涉及到比较两个序列的元素是否相等,然后根据不同的情况进行状态转移。 对于区间长度为1的情况,可以先进行初始化,然后再通过枚举区间长度和区间左端点,计算出dp[i][j]的值。 以下是一个示例代码,展示了如何使用区间DP来解决一个字符串匹配的问题: #include <cstdio> #include <cstring> #include <string> #include <iostream> #include <algorithm> using namespace std; const int maxn=510; const int inf=0x3f3f3f3f; int n,dp[maxn][maxn]; char s[maxn]; int main() { scanf("%d", &n); scanf("%s", s + 1); for(int i = 1; i <= n; i++) dp[i][i] = 1; for(int i = 1; i <= n; i++) { if(s[i] == s[i - 1]) dp[i][i - 1] = 1; else dp[i][i - 1] = 2; } for(int len = 3; len <= n; len++) { int r; for(int l = 1; l + len - 1 <= n; l++) { r = l + len - 1; dp[l][r] = inf; if(s[l] == s[r]) dp[l][r] = min(dp[l + 1][r], dp[l][r - 1]); else { for(int k = l; k <= r; k++) { dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]); } } } } printf("%d\n", dp[n]); return 0; } 希望这个例子能帮助你理解区间DP的基本思想和应用方法。如果你还有其他问题,请随时提问。
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