【codeforces 701C】They Are Everywhere

【题目链接】:http://codeforces.com/contest/701/problem/C

【题意】



让你选择一段最短的区间;

使得这段区间里面包含所有种类的字符;

【题解】



之前都是用二分写;

现在会用类似队列的思路写了;

就是先确定左端点;

然后右端点右移;

直到出现所有种类;(这时候右端点就没必要再右移了)

然后右端点不动,右移左端点;

然后如果这时候又没有全部种类,就再右移右端点;

O(N)的复杂度吧



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define ps push_back

#define fi first

#define se second

#define rei(x) cin >> x

#define pri(x) cout << x

#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 1e5+100;

map <char,int> dic;

int n,tot,ans=N;

char s[N];

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    ios::sync_with_stdio(false);

    rei(n);

    rei((s+1));

    rep1(i,1,n)

        if (!dic[s[i]])

        {

            dic[s[i]] = 1;

            tot++;

        }

    dic.clear();

    int l = 1,r = 1,now = 0;

    dic[s[1]] = 1;

    now = 1;

    while (r<=n)

    {

        if (l<=r && now==tot)

        {

            ans = min(ans,r-l+1);

            dic[s[l]]--;

            if (dic[s[l]]==0) now--;

            l++;

        }

        else

        {

            r++;

            if (r<=n)

            {

                dic[s[r]]++;

                if (dic[s[r]]==1) now++;

            }

        }

    }

    pri(ans<<endl);

    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);

    return 0;

}