Codeforces Round #426 (Div. 2) B. The Festive Evening

B. The Festive Evening
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output

It’s the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it’s a necessity to not let any uninvited guests in.

There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest’s arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.

For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are k such guards in the castle, so if there are more than k opened doors, one of them is going to be left unguarded! Notice that a guard can’t leave his post until the door he is assigned to is closed.

Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than k doors were opened.

Input
Two integers are given in the first string: the number of guests n and the number of guards k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26).

In the second string, n uppercase English letters s1s2… sn are given, where si is the entrance used by the i-th guest.

Output
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.

You can output each letter in arbitrary case (upper or lower).

Examples
input
5 1
AABBB
output
NO
input
5 1
ABABB
output
YES
Note
In the first sample case, the door A is opened right before the first guest’s arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.

In the second sample case, the door B is opened before the second guest’s arrival, but the only guard can’t leave the door A unattended, as there is still one more guest that should enter the castle through this door.

题意：一个房子有A-Z个门，来了A人只能进A门，B人只能进B门，依次类推。
给出了n个人来的顺序，还有k个警卫，一个门至少站一个警卫，直到这个门的最后一个人进入，警卫才能离开这个门，如果开的门的数量大于警卫的数量输出“Yes”，否则输出“No”。
做法：每个门第一个人到的时候+1，最后一个人到的时候-1。
但是直接这样写会错，比如：
27 1
ABCDEFGHIJKLMNOPQRSTUVWXYZA
YES

因为最后一个和第一个是同一个位置，就相当于没加没减。
a[1000005].//每个字母对应的位置
b[100]。//每个字母出现的次数
string s;//进去人的顺序
那么可以换一种思路，记录这个门总共进去多少人，进去的时候记录一下，如果b[a[i]]==0，a[i]++,然后b[a[i]]++；再来一遍循环找最后一个进去的人，找到那个最后一个进去的人，就a[i+1]–,这样能避免前面那个错误。

#include<stdio.h>#include<string>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int a[1000005];int b[100];//代表字母第几次出现int main(){    memset(b,0,sizeof(b));    memset(a,0,sizeof(a));    int n,k;    string s;    scanf("%d%d",&n,&k);    cin>>s;    for(int i=0; i<n; i++)    {        if(b[s[i]]==0)            a[i]++;        b[s[i]]++;    }    for(int i=0; i<n; i++)    {        b[s[i]]--;        if(b[s[i]]==0)            a[i+1]--;//自己的门守完后再去别的门守。    }    int sum=0;    int maxx=0;    for(int i=0; i<n; i++)    {        sum+=a[i];        maxx=max(maxx,sum);    }    if(maxx>k)        printf("YES\n");    else        printf("NO\n");    return 0;}