Codeforces Round #364 (Div. 2)C. They Are Everywhere(尺取法)
題目鏈接:
C. They Are Everywhere
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard outputSergei B., the young coach of Pokemons, has found the big house which consists of?n?flats ordered in a row from left to right. It is possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number?1?is only connected with the flat number?2?and the flat number?n?is only connected with the flat number?n?-?1.
There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street, visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.
Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.
InputThe first line contains the integer?n?(1?≤?n?≤?100?000) — the number of flats in the house.
The second line contains the row?s?with the length?n, it consists of uppercase and lowercase letters of English alphabet, the?i-th letter equals the type of Pokemon, which is in the flat number?i.
OutputPrint the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.
Examples input 3AaA output 2 input 7
bcAAcbc output 3 input 6
aaBCCe output 5 Note
In the first test Sergei B. can begin, for example, from the flat number?1?and end in the flat number?2.
In the second test Sergei B. can begin, for example, from the flat number?4?and end in the flat number?6.
In the third test Sergei B. must begin from the flat number?2?and end in the flat number?6.
?這次沒花過長(zhǎng)時(shí)間看出尺取法來做,在尺取法的變形上多想了想。
首先預(yù)處理出總共有多少類型。
然后就普通的尺取法了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <map> 6 using namespace std; 7 typedef long long ll; 8 const int maxn = 1e5+5; 9 char s[maxn]; 10 map<char,int> mp,mp1; 11 int main() 12 { 13 int n; 14 scanf("%d",&n); 15 scanf("%s",s+1); 16 int type = 0; 17 for(int i=1;i<=n;i++) 18 { 19 if(!mp.count(s[i])) 20 { 21 mp1[s[i]] = 1; 22 type++; 23 mp[s[i]] = 0; 24 } 25 } 26 int sum = 0; 27 int t = 1; 28 int ans = 1e9+7; 29 for(int l=1;l<=n;l++) 30 { 31 while(t<=n&&sum<type) 32 { 33 if(mp[s[t]]==0) 34 { 35 sum++; 36 } 37 mp[s[t]]++; 38 t++; 39 } 40 if(sum<type) break; 41 ans = min(ans,t-l); 42 mp[s[l]]--; 43 if(mp[s[l]]==0) 44 { 45 sum--; 46 } 47 } 48 printf("%d\n",ans); 49 return 0; 50 }?
轉(zhuǎn)載于:https://www.cnblogs.com/littlepear/p/5815930.html
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