CF580B Kefa and Company（滑动窗口）

题目描述

Sergei 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.

输入格式

The first line contains the integer n ( 1<=n<=100000 ) — 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 .

输出格式

Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.

隐藏翻译

题意翻译

给出一个长度为n的序列，求出长度最小的序列that包含序列中所有的字母（区别大小写）。详见样例

输入输出样例

输入 #1复制

3
AaA

输出 #1复制

2

输入 #2复制

7
bcAAcbc

输出 #2复制

3

输入 #3复制

6
aaBCCe

输出 #3复制

5

说明/提示

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 .

#include<bits/stdc++.h>
#define int long long
#define ll long long
#define N 100005
#define INF INT64_MAX
#define MINF 0x3f
#define eps 1e-9
using namespace std;
int n,x,m,k,t,ans=0,vis[N],c[N],minn=INF,maxn=-INF,flag=false;
int a[N],b[N],t1[N],t2[N],sum[N];
map<int,int> mp;  //map 数组作映射，记录前缀和相同的时候
struct node{    //因为要对a数组进行排序，所以用结构体排序
  int a,b;
}no[N];

bool cmp(node p,node q)
{
  return p.a<q.a;
}

void solve()
{
  cin>>n>>m;
  for(int i=1;i<=n;i++) {
    cin>>no[i].a>>no[i].b; 
  }
sort(no+1,no+n+1,cmp);
  for(int i=1;i<=n;i++) 
  {
    sum[i]=sum[i-1]+no[i].b;   //前缀和计算b的前i项和
  }
  int r=1;
  for(int l=1;l<=n;l++)         //滑动窗口，缩小左边界
  {
    while(r<=n&&no[r].a-no[l].a<m) //当右边界不越界，且ai差距不超过m的时候
    {
      ans=max(ans,sum[r]-sum[l-1]);  //就更新ans的长度，寻找的在此范围内的最大ans
      r++;                   //找完以后右边界继续右移
    }
  }

}
signed main()
{
  
  ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
     solve();
    return 0; 
 }