hdu 2852 KiKi's K-Number(线段树单点更新）
姚煜
题目：http://acm.hdu.edu.cn/showproblem.php?pid=2852
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
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
Sample Output
No Elment! 6 Not Find! 2 2 4 Not Find!
WA：
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=1e5;
struct node{
int l,r,count;
int mid(){ return (l+r)/2; }
}tree[maxn<<2];
void build(int root,int a,int b){
tree[root].l=a;
tree[root].r=b;
tree[root].count=0;
if(a==b) return ;
int m=tree[root].mid();
build(2*root,a,m);
build(2*root+1,m+1,b);
}
void insert(int root,int a,int b,int number){
if(a==b){
tree[root].count++;
return ;
}
int m=(a+b)/2;
if(number<=m)insert(2*root,a,m,number);
else if(number>m)insert(2*root+1,m+1,b,number);
tree[root].count=tree[2*root].count+tree[2*root+1].count;
}
bool work;
void delet(int root,int a,int b,int number){
if(tree[root].count<1){
work=0;
return ;
}
if(a==b){
tree[root].count--;
return ;
}
int m=(a+b)/2;
if(number<=m)delet(2*root,a,m,number);
else if(number>m)delet(2*root+1,m+1,b,number);
tree[root].count=tree[2*root].count+tree[2*root+1].count;
}
int findres;
void inquiry(int root,int a,int b,int q1,int q2){
if(tree[root].count<1){
printf("Not Find!\n");
return ;
}
if(a==b){
printf("%d\n",a);
return ;
}
int m=(a+b)/2;
if(q1<=m&&q2<=tree[2*root].count)
inquiry(2*root,a,m,q1,q2);
else {
if(q1>=tree[2*root].r)q2=q2-tree[2*root].count;
inquiry(2*root+1,m+1,b,q1,q2);
}
}
int main()
{
//freopen("cin.txt","r",stdin);
int m,p,a,b;
while(cin>>m){
build(1,1,maxn);
while(m--){
scanf("%d",&p);
if(p==0){
scanf("%d",&a);
insert(1,1,maxn,a);
}
else if(p==1){
scanf("%d",&a);
work=1;
delet(1,1,maxn,a);
if(work==0){
printf("No Elment!\n");
}
}
else {
scanf("%d%d",&a,&b);
inquiry(1,1,maxn,a,b);
}
}
}
return 0;
}
AC：
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=1e5;
struct node{
int l,r,count;
int mid(){ return (l+r)/2; }
}tree[maxn<<4];
void build(int root,int a,int b){
tree[root].l=a;
tree[root].r=b;
tree[root].count=0;
if(a==b) return ;
int m=tree[root].mid();
build(2*root,a,m);
build(2*root+1,m+1,b);
}
void update(int root,int l,int r,int number,int val){
if(l==r){
tree[root].count+=val;
return ;
}
int m=tree[root].mid();
if(number<=m)update(2*root,l,m,number,val);
else update(2*root+1,m+1,r,number,val);
tree[root].count=tree[2*root].count+tree[2*root+1].count;
}
int sum(int root,int l,int r,int ql,int qr){ // l and r is necessary
if(qr<l||ql>r){
return 0;
}
if(ql<=l&&qr>=r){ //care >= == <=
return tree[root].count;
}
int m=(l+r)/2;
return sum(2*root,l,m,ql,qr)+sum(2*root+1,m+1,r,ql,qr);
}
int findgoal(int root,int l,int r,int number){
if(l==r){
return tree[root].l;
}
int m=(l+r)/2;
if(number>tree[2*root].count){
//if(l==r){
// return tree[root].l;
//}
return findgoal(2*root+1,m+1,r,number-tree[2*root].count);
}
return findgoal(2*root,l,m,number);
}
int main()
{
//freopen("cin.txt","r",stdin);
int m,p,a,b;
while(cin>>m){
build(1,1,maxn);
while(m--){
scanf("%d",&p);
if(p==0){
scanf("%d",&a);
update(1,1,maxn,a,1);
//for(int i=20;i>0;i--)cout<<tree[i].count<<" "; cout<<endl;
}
else if(p==1){
scanf("%d",&a);
if(sum(1,1,maxn,1,a)==sum(1,1,maxn,1,a-1)){
printf("No Elment!\n");
}
else update(1,1,maxn,a,-1);
}
else {
scanf("%d%d",&a,&b);
int goal=sum(1,1,maxn,1,a)+b;
//cout<<"goal: "<<goal<<endl;
int ans=findgoal(1,1,maxn,goal);
//cout<<"ans: "<<ans<<endl;
if(ans==maxn)printf("Not Find!\n");
else printf("%d\n",ans);
}
}
}
return 0;
}还有一份长得很像的 Runtime Error，最开始我真没看出来哪里不同，后来我发现tree[maxn<<4]; 如果写成tree[maxn<<2]; 或者：tree[maxn*5] 居然会RE，我又惊呆了。。。
类似资料：