KiKi's K-Number
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3966 Accepted Submission(s): 1758
Problem Description
For the k-th number, we all should be very familiar with it. Of course,to kiki it is also simple. Now Kiki meets a very similar problem, kiki wants to design a container, the container is to support the three operations.
Push: Push a given element e to container
Pop: Pop element of a given e from container
Query: Given two elements a and k, query the kth larger number which greater than a in container;
Although Kiki is very intelligent, she can not think of how to do it, can you help her to solve this problem?
Input
Input some groups of test data ,each test data the first number is an integer m (1 <= m <100000), means that the number of operation to do. The next m lines, each line will be an integer p at the beginning, p which has three values:
If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container.
If p is 0, then there will be an integer e (0 <e <100000), means press element e into Container.
If p is 1, then there will be an integer e (0 <e <100000), indicated that delete the element e from the container
If p is 2, then there will be two integers a and k (0 <a <100000, 0 <k <10000),means the inquiries, the element is greater than a, and the k-th larger number.
Output
For each deletion, if you want to delete the element which does not exist, the output "No Elment!". For each query, output the suitable answers in line .if the number does not exist, the output "Not Find!".
Sample Input
5
0 5
1 2
0 6
2 3 2
2 8 1
7
0 2
0 2
0 4
2 1 1
2 1 2
2 1 3
2 1 4
0 5
1 2
0 6
2 3 2
2 8 1
7
0 2
0 2
0 4
2 1 1
2 1 2
2 1 3
2 1 4
Sample Output
No Elment!
6
Not Find!
2
2
4
Not Find!
6
Not Find!
2
2
4
Not Find!
Source
题意：
一个空的容器，进行三种操作：
0 a ： 把a加进容器里；
1 a ： 把a从容器中删除，如果没有a则输出No Elment!,如果有多个a则只删除一个；
2 a k ： 求整个容器中所有比a大的数中的第k大的数，并输出，如果没有则输出NO Finds！。
题解：因为数的范围比较小 所以不需要离散化。主席树记录线段树的历史版本。查找容器中比a大的第k大的数字可以 先查询范围在[1,a]的数字数量q
求第q+k大即可。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <map>
#define ll __int64
#define mod 1000000007
#define dazhi 2147483647
const int N=;
using namespace std;
struct chairmantree
{
int rt[*N],ls[*N],rs[*N],sum[*N];
int tot;
void init()
{
tot=;
}
void build(int l,int r,int &pos)
{
pos=++tot;
sum[pos]=;
if(l==r) return ;
int mid=(l+r)>>;
build(l,mid,ls[pos]);
build(mid+,r,rs[pos]);
}
void update(int p,int c,int l,int r,int pre,int &pos)
{
pos=++tot;
ls[pos]=ls[pre];
rs[pos]=rs[pre];
sum[pos]=sum[pre]+c;
if(l==r) return ;
int mid=(l+r)>>;
if(p<=mid)
update(p,c,l,mid,ls[pre],ls[pos]);
else
update(p,c,mid+,r,rs[pre],rs[pos]);
}
int rank(int s,int e,int l,int r,int L,int R)
{
if(s<=l&&e>=r) return sum[R]-sum[L];
int ans=;
int mid=(l+r)>>;
if(s<=mid)
ans+=rank(s,e,l,mid,ls[L],ls[R]);
if(e>mid)
ans+=rank(s,e,mid+,r,rs[L],rs[R]);
return ans;
}
int query(int L,int R,int l,int r,int k)
{
if(l==r) return l;
int mid=(l+r)>>;
int x=sum[ls[R]]-sum[ls[L]];
if(k<=x)
query(ls[L],ls[R],l,mid,k);
else
query(rs[L],rs[R],mid+,r,k-x);
}
} tree;
int m;
int exm;
int e;
int main()
{
int rr=;
while(scanf("%d",&m)!=EOF)
{
tree.init();
tree.build(,rr,tree.rt[]);
for(int i=; i<=m; i++)
{
scanf("%d",&exm);
if(exm==)
{
scanf("%d",&e);
tree.update(e,,,rr,tree.rt[i-],tree.rt[i]);
}
if(exm==)
{
scanf("%d",&e);
if(tree.rank(e,e,,rr,tree.rs[],tree.rt[i-]))
{
tree.update(e,-,,rr,tree.rt[i-],tree.rt[i]);
}
else
{
tree.update(e,,,rr,tree.rt[i-],tree.rt[i]);
printf("No Elment!\n");
}
}
if(exm==)
{
int a,k;
scanf("%d %d",&a,&k);
tree.update(,,,rr,tree.rt[i-],tree.rt[i]);
int q=tree.rank(,a,,rr,tree.rt[],tree.rt[i]);
int xx=tree.rank(,rr,,rr,tree.rt[],tree.rt[i]);
k+=q;
if(k>xx)
printf("Not Find!\n");
else
printf("%d\n",tree.query(tree.rt[],tree.rt[i],,rr,k));
}
}
}
return ;
}