No to Palindromes! CodeForces - 464A

原题链接
考察:贪心(?)
思路:
  不存在回文子串的条件是对于任意\(i\),均有:

\[s[i]!=s[i-1] , s[i] != s[i-2] \]

  DFS枚举每一个修改的值.

Code

#include <iostream>
#include <cstring>
using namespace std;
const int N = 1010;
char s[N],b[N];
int n,p;
bool dfs(int u,bool limit)
{
	if(u>n)
	{
		memcpy(b,s,sizeof s);
		return 1;
	}
	char t = s[u];
	char st = limit?s[u]:'a';
	if(u==n&&st==t) st++; 
	for(char c=st;c<='a'+p-1;c++)
	{
		s[u] = c;
		if(u>=1&&s[u-1]==s[u]) continue;
		if(u>=2&&s[u-2]==s[u]) continue;
		if(dfs(u+1,limit&&c==t)) return 1;
		s[u] = t;
	}
	return 0;
}
int main()
{
	scanf("%d%d%s",&n,&p,s+1);
	if(dfs(1,1)) printf("%s\n",b+1);
	else puts("NO");
	return 0;
}