HDOJ 题目2852 KiKi's K-Number(线段树求大于a的第k值)
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KiKi's K-Number
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3258 Accepted Submission(s): 1459
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?
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 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.
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
50 51 20 62 3 22 8 170 20 20 42 1 12 1 22 1 32 1 4
Sample Output
No Elment!6Not Find!224Not Find!
Source
2009 Multi-University Training Contest 4 - Host by HDU
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题目大意:三种操作0表示加入一个数,1表示删除一个数,2表示查询比a大的第k个数
153845082015-11-04 21:12:23Accepted28521014MS3120K2695 BC++XY_
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#define n 100001#define LL long longusing namespace std;int node[n<<2];int vis[n];void pushup(int tr){ node[tr]=node[tr<<1]+node[tr<<1|1];}void build(int l,int r,int tr){ node[tr]=0; if(l==r) { return; } int mid=(l+r)>>1; build(l,mid,tr<<1); build(mid+1,r,tr<<1|1);}void update(int pos,int val,int l,int r,int tr){ if(l==r) { node[tr]+=val; return; } int mid=(l+r)>>1; if(pos<=mid) { update(pos,val,l,mid,tr<<1); } else update(pos,val,mid+1,r,tr<<1|1); pushup(tr);}int query(int L,int R,int l,int r,int tr){ if(L==l&&r==R) { return node[tr]; } int mid=(l+r)>>1; if(R<=mid) return query(L,R,l,mid,tr<<1); else if(L>mid) return query(L,R,mid+1,r,tr<<1|1); else { int a=query(L,mid,l,mid,tr<<1); int b=query(mid+1,R,mid+1,r,tr<<1|1); return a+b; }}int find(int num,int l,int r,int tr){ if(node[tr]<num) return -1; if(l==r) return l; int mid=(l+r)>>1; if(node[tr<<1]>=num) find(num,l,mid,tr<<1); else find(num-node[tr<<1],mid+1,r,tr<<1|1);}int main(){ int q; while(scanf("%d",&q)!=EOF) { build(1,n,1); memset(vis,0,sizeof(vis)); while(q--) { int op; scanf("%d",&op); if(op==0) { int x; scanf("%d",&x); update(x,1,1,n,1); vis[x]++; } else { if(op==1) { int x; scanf("%d",&x); if(vis[x]==0) printf("No Elment!\n"); else { update(x,-1,1,n,1); vis[x]--; } } else { int a,b; scanf("%d%d",&a,&b); int num=query(1,a,1,n,1); // printf("+++++%d\n",num); num+=b; // printf("++++++%d\n",num); int c=find(num,1,n,1); if(c==-1) printf("Not Find!\n"); else printf("%d\n",c); } } } }}
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