HUAS 1483 mex(莫队算法)
考虑莫队算法,对于区间减小的情况,可以O(1)解决。对于区间增加的情况,可能需要O(n)解决。好在数据不卡莫队。
1200ms过了。
离线+线段树 760ms过了。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 1000000009 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=200005; //Code begin... struct Node{int l, r, l1, id;}node[N]; int a[N], ans[N], unit, q, num[N]; bool comp(Node a, Node b){ if (a.l1!=b.l1) return a.l1<b.l1; return a.r<b.r; } void sol(){ int tmp=0, l=1, r=0; FOR(i,1,q) { while (r<node[i].r) { ++r; ++num[a[r]]; if (a[r]!=tmp) continue; for (int now=tmp+1; ; ++now) if (!num[now]) {tmp=now; break;} } while (r>node[i].r) { --num[a[r]]; if (a[r]<tmp&&!num[a[r]]) tmp=a[r]; --r; } while (l<node[i].l) { --num[a[l]]; if (a[l]<tmp&&!num[a[l]]) tmp=a[l]; ++l; } while (l>node[i].l) { --l; ++num[a[l]]; if (a[l]!=tmp) continue; for (int now=tmp+1; ; ++now) if (!num[now]) {tmp=now; break;} } ans[node[i].id]=tmp; } } int main () { int n; n=Scan(); q=Scan(); unit=(int)sqrt(n); FOR(i,1,n) a[i]=Scan(); FOR(i,1,q) node[i].l=Scan(), node[i].r=Scan(), node[i].id=i, node[i].l1=node[i].l/unit; sort(node+1,node+q+1,comp); sol(); FOR(i,1,q) Out(ans[i]), putchar('\n'); return 0; }