【codeforces 701C】They Are Everywhere-蒲公英云

【codeforces 701C】They Are Everywhere

小鱼儿 2021-10-01 07:08 273阅读 0赞

【题目链接】//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】

【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;
}