hdu 2158 最短区间版大家来找碴(尺取法)

#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int n,m,q;
int a[100007];        //询问数组 
int temp[107];
bool vis[100007]; //标记为需求数 
int num[100007]; //需求数的数量 
int main(){
    ios::sync_with_stdio(false);
    while(~scanf("%d%d",&n,&m)){
        if(!n&&!m) break;
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        for(int i=1;i<=m;i++){
            memset(vis,0,sizeof(vis));
            memset(num,0,sizeof(num));
            scanf("%d",&q);
            int sum=0;
            for(int j=1;j<=q;j++){
                scanf("%d",temp+j);
                if(!vis[temp[j]]) vis[temp[j]]=1,sum++;
            }
            int ans=inf; int u,v,s;
            u=v=1;
            s=0;
            while(1){
                while(v<=n&&s<sum){
                    if(vis[a[v]]&&!num[a[v]]) num[a[v]]++,s++; //如果是需求数且没出现则s++ 
                    else if(num[a[v]]) num[a[v]]++;//不然只记录下数目 
                    v++;
                }
                if(s<sum) break;
                ans=min(ans,v-u);
                if(vis[a[u]]&&num[a[u]]==1) num[a[u]]--,s--;    //与上面对应 
                else if(num[a[u]]>1) num[a[u]]--;
                u++;
            }
            cout<<ans<<endl;
        }
    }
}