Codeforces 464A. No to Palindromes! (dp,向前推理,向后查询)

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本文探讨如何在给定长度和字母限制下，找到一个符合特定条件的字符串的下一个可能选项，即该字符串不能包含长度为2或更长的回文子串，并且每个字符仅限于字母表前p个字母。

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Paul hates palindromes. He assumes that string s is tolerable if each its character is one of the first p letters of the English alphabet and sdoesn't contain any palindrome contiguous substring of length 2 or more.

Paul has found a tolerable string s of length n. Help him find the lexicographically next tolerable string of the same length or else state that such string does not exist.

Input

The first line contains two space-separated integers: n and p (1 ≤ n ≤ 1000; 1 ≤ p ≤ 26). The second line contains string s, consisting ofn small English letters. It is guaranteed that the string is tolerable (according to the above definition).

Output

If the lexicographically next tolerable string of the same length exists, print it. Otherwise, print "NO" (without the quotes).

Sample test(s)

input
3 3
cba

output
NO

input
3 4
cba

output
cbd

input
4 4
abcd

output
abda

Note

String s is lexicographically larger (or simply larger) than string t with the same length, if there is number i, such that s1 = t1, ..., si = ti, si + 1 > ti + 1.

The lexicographically next tolerable string is the lexicographically minimum tolerable string which is larger than the given one.

A palindrome is a string that reads the same forward or reversed.

http://codeforces.com/contest/464/problem/A

//Hello. I'm Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi 3.1415926535898
#define eps 1e-6
#define MOD 1000000007
#define MAXN 100100
#define M 200
int n,p;
char c,to;
string row,ans;
bool is_true_ok(string s,int pos)
{
    bool ok,is_ok=true;
    int len=slen(s);
    rep(i,pos+1,len)
    {
        ok=false;
        for(c='a';c<=to;c++)
        {
            if(i==0 || (i>0&&c!=s[i-1]))
            {
                if(i==1 || (i>1&&c!=s[i-2]))
                {
                    ok=true;
                    s[i]=c;
                    break;
                }
            }
        }
        if(!ok)
        {
            is_ok=false;
            break;
        }
    }
    if(!is_ok) return false;
    else
    {
        ans=s;
        return true;
    }
}
int main()
{
//    input;
    string s;
    bool ok;
    cin>>n>>p;
    to=(p-1)+'a';
    cin>>row;
    depin(i,n-1,0)
    {
        ok=false;
        for(c=row[i]+1;c<=to;c++)
        {
            if(i==0 || (i>0&&c!=row[i-1]))
            {
                if(i==0 || i==1 || (i>1&&c!=row[i-2]))
                {
                    ok=true;
                    break;
                }
            }
        }
        if(ok)
        {
            s=row;
            s[i]=c;
            if(is_true_ok(s,i))
            {
                cout<<ans<<endl;
                exit(0);
            }
        }
    }
    cout<<"NO"<<endl;
}