Codeforces 196 D. The Next Good String,codeforces196
D. The Next Good Stringtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output
In problems on strings one often has to find a string with some particular properties. The problem authors were reluctant to waste time on thinking of a name for some string so they called it good. A string is good if it doesn't have palindrome substrings longer than or equal to d.
You are given string s, consisting only of lowercase English letters. Find a good string t with length |s|, consisting of lowercase English letters, which is lexicographically larger than s. Of all such strings string t must be lexicographically minimum.
We will call a non-empty string s[a ... b] = sasa + 1... sb (1 ≤ a ≤ b ≤ |s|) a substring of string s = s1s2... s|s|.
A non-empty string s = s1s2... sn is called a palindrome if for all i from 1 to n the following fulfills: si = sn - i + 1. In other words, palindrome read the same in both directions.
String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there exists such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1. Characters in such strings are compared like their ASCII codes.
Input
The first line contains integer d (1 ≤ d ≤ |s|).
The second line contains a non-empty string s, its length is no more than 4·105 characters. The string consists of lowercase English letters.
Output
Print the good string that lexicographically follows s, has the same length and consists of only lowercase English letters. If such string does not exist, print "Impossible" (without the quotes).
Sample test(s)input
3aaaaaaa
output
aabbcaa
input
3zzyzzzz
output
Impossible
input
4abbabbbabbb
output
abbbcaaabab
搜尋+hash,
長度為L的迴文串一定包含長度為L-2的迴文串。
判斷迴文的時候只要判斷d和d+1不是迴文串就可以了。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef unsigned long long int ull;const int maxn=500100;char s[maxn],r[maxn];int n,d;ull p[maxn],hash[maxn],rhash[maxn];bool ok(int ed,int d){ed++;int st=ed-d+1;if(st<0) return true;if((rhash[ed]-rhash[st-1]*p[d])*p[st-1]!=hash[ed]-hash[st-1])return true;return false;}bool dfs(int x,int t){if(x==n){puts(r);return true;}for(r[x]=(t?s[x]:'a');r[x]<='z';r[x]++){hash[x+1]=hash[x]+r[x]*p[x];rhash[x+1]=rhash[x]*175+r[x];if(ok(x,d)&&ok(x,d+1)&&dfs(x+1,t&&(r[x]==s[x])))return true;}return false;}int main(){scanf("%d %s",&d,s);n=strlen(s);int i=n-1;for(;i>=0&&s[i]=='z';i--)s[i]='a';if(i<0){puts("Impossible");return 0;}s[i]++;p[0]=1;for(int i=1;i<n+100;i++)p[i]=p[i-1]*175;if(dfs(0,1)==false)puts("Impossible");return 0;}