按題目給的tips,s+=s;之後從頭開始周遊,找到所有的r...w...b結構,記錄首尾兩個offset然後向左右周遊,遇到w或者與對應的offset處一樣的話長度就++;
坑:
最後結果如果大于n說明有重複
如果為0說明全為w
防止漏掉結尾一段,手動給結尾處一個斷
代碼:
變量解釋:
n長度;s字元串;tc儲存周遊時目前的顔色;of1,of2前後兩個偏移;tof1,tof2臨時;almax最長長度;tmax臨時長度;
/*
ID: windroid
LANG: C++
TASK: beads
*/
#include<iostream>
#include<fstream>
#include<string>
using namespace std;
int main(){
ifstream fin("beads.in");
ofstream fout("beads.out");
int n;
string s;
fin>>n>>s;
//cin>>s;
s+=s;
char tc=0;
int of1=0,of2=0;
int almax=0;
for(int i=0;i<2*n;i++){
if(tc==0&&s[i]!='w'){
tc=s[i];
of1=i;
}
if(s[i]!='w'&&s[i]!=tc||i==2*n-1){
// cout<<"|"<<i<<"|";
//cout<<"|";
if(of2!=0){
of1=of2;
}
tc=s[i];
of2=i;
int tmax=of2-of1+1;
int tof1=of1,tof2=of2;
while(tof1>0&&(s[tof1-1]=='w'||s[tof1-1]==s[of1])){
tmax++;
tof1--;
}
while(tof2<2*n-1&&(s[tof2+1]=='w'||s[tof2+1]==s[of2])){
tmax++;
tof2++;
}
// if(tmax==73) cout<<tof1<<"=="<<tof2<<endl;
if(tmax>almax){
almax=tmax;
}
}
// cout<<s[i];
}
//cout<<endl<<almax<<endl;
if(almax==0||almax>=n) almax=n;
fout<<almax<<endl;
return 0;
}