3685: 普通van Emde Boas树
Time Limit: 9 Sec Memory Limit: 128 MBSubmit: 1932 Solved: 626
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Description
设计数据结构支持:
1 x 若x不存在,插入x
2 x 若x存在,删除x
3 输出当前最小值,若不存在输出-1
4 输出当前最大值,若不存在输出-1
5 x 输出x的前驱,若不存在输出-1
6 x 输出x的后继,若不存在输出-1
7 x 若x存在,输出1,否则输出-1
Input
第一行给出n,m 表示出现数的范围和操作个数
接下来m行给出操作
n<=10^6,m<=2*10^6,0<=x<n
Output
Sample Input
10 11
1 1
1 2
1 3
7 1
7 4
2 1
3
2 3
4
5 3
6 2
Sample Output
1
-1
2
2
2
-1
HINT
Source
By Zky
常规操作
要用zkw减小常数,但我貌似卡过了
#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#define ls u<<1#define rs ls|1#define ll long long#define N 1000050using namespace std;int n,m,sum[N<<2],vis[N];void pushup(int u){sum[u]=sum[ls]+sum[rs];}int findpos(int u,int l,int r,int x){if(!sum[u])return 0;if(l==r)return sum[u];int mid=l+r>>1;if(x<=mid)return findpos(ls,l,mid,x);return findpos(rs,mid+1,r,x)+sum[ls];}void update(int u,int L,int R,int p,int val){if(L==R){sum[u]=val;return;}int mid=L+R>>1;if(p<=mid)update(ls,L,mid,p,val);else update(rs,mid+1,R,p,val);pushup(u);}int query(int u,int l,int r,int p){if(l==r)return l;int mid=l+r>>1;if(p<=sum[ls])return query(ls,l,mid,p);return query(rs,mid+1,r,p-sum[ls]); }int main(){scanf("%d%d",&n,&m);int op,x;while(m--){scanf("%d",&op);if(op!=3&&op!=4)scanf("%d",&x);if(op==1)update(1,0,n-1,x,1),vis[x]=1;if(op==2)update(1,0,n-1,x,0),vis[x]=0;if(op==3){if(sum[1]==0)puts("-1");else printf("%d\n",query(1,0,n-1,1));}if(op==4){if(sum[1]==0)puts("-1");else printf("%d\n",query(1,0,n-1,sum[1]));}if(op==5){int p=findpos(1,0,n-1,x);if(vis[x])p--;if(p<=0)puts("-1");else printf("%d\n",query(1,0,n-1,p));}if(op==6){int p=findpos(1,0,n-1,x);p++;if(p>sum[1])puts("-1");else printf("%d\n",query(1,0,n-1,p));}if(op==7){if(vis[x])puts("1");else puts("-1");}}return 0;}
