bzoj3585: mex

来源：互联网发布：手机惯性导航算法编辑：程序博客网时间：2024/04/25 02:11

传送门：http://www.lydsy.com:808/JudgeOnline/problem.php?id=3585

思路：又是一道莫队+分块。

对权值分块，大于n的直接不用管。

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>const int maxn=200010,maxb=450;using namespace std;struct que{int l,r,id;}q[maxn];int n,m,sz,cnt,a[maxn],s[maxn],bel[maxn],l[maxb],r[maxb],bloans[maxn],ans[maxn];bool cmp(que a,que b){return bel[a.l]==bel[b.l]?a.r<b.r:bel[a.l]<bel[b.l];}void add(int x){if (x>n) return;if (!s[x]++) bloans[bel[x]]++;}void del(int x){if (x>n) return;if (!--s[x]) bloans[bel[x]]--;}int query(){int i;if (!s[0]) return 0;for (i=1;i<=cnt;i++)if (bloans[i]!=r[i]-l[i]+1) break;for (int j=l[i];j<=r[i];j++)if (!s[j]) return j;}void work(){for (int i=1,l=1,r=0;i<=m;i++){for (;r<q[i].r;r++) add(a[r+1]);for (;r>q[i].r;r--) del(a[r]);for (;l<q[i].l;l++) del(a[l]);for (;l>q[i].l;l--) add(a[l-1]);ans[q[i].id]=query();}for (int i=1;i<=m;i++) printf("%d\n",ans[i]);}int main(){scanf("%d%d",&n,&m),sz=sqrt(n),cnt=n/sz+(n%sz>0);for (int i=1;i<=n;i++) bel[i]=(i-1)/sz+1;for (int i=1;i<=n;i++){r[bel[i]]=i;if (!l[bel[i]]) l[bel[i]]=i;}for (int i=1;i<=n;i++) scanf("%d",&a[i]),a[i];for (int i=1;i<=m;i++) scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i;sort(q+1,q+1+m,cmp),work();return 0;}

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