The kth great number（第k大数模板题：优先队列或树状数组或SBT）
Link：http://acm.hdu.edu.cn/showproblem.php?pid=4006
The kth great numberTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 7134 Accepted Submission(s): 2916
Problem Description
Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.
Input
There are several test cases. For each test case, the first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number.
Output
The output consists of one integer representing the largest number of islands that all lie on one line.
Sample Input
Sample Output
Source
The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest
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题意:题目会给出n个数,求第k大的数.
输入:第一行输入两个整数n 和 k
接下来有n行数据,输入的数据分为两种.
输入I 和 一个数字x 表示写入数字x
输入Q 表示进行一次询问 询问当前第k大的数是多少并输出这个数.
开始使用sort进行排序,提交后会超时.
后来改用优先队列.
声明一个结构体
struct Node
{
int x;
friend bool operator < (Node a,Node b)
{
return a.x > b.x;
}
};
主要用于重载小于号,让小的优先.
输入时先输入k个数,输入k个数后再输入时需要判断输入的数与队头的大小关系.
输入数小于队头则不进队.
输入数大于队头则进队,弹出队头.
保证优先队列中只有k的元素.
而队头是k个元素中最小的元素.
即第k大元素
AC code：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
struct node{
int v;
friend bool operator < (node a,node b)
{
return a.v>b.v;
}
};
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
int cnt=0;
priority_queue<node>pq;
while(n--)
{
char c;
cin>>c;
if(c=='I')
{
if(cnt<k)
{
node t;
scanf("%d",&t.v);
pq.push(t);
cnt++;
}
else
{
node t;
scanf("%d",&t.v);
if(pq.top().v<t.v)
{
pq.pop();
pq.push(t);
}
}
}
else
{
if(c=='Q')
{
printf("%d\n",pq.top().v);
}
}
}
}
return 0;
}
以下树状数组代码参考：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;
const int N=1000005;
int cal[N],ptr[N];
struct TNode {
char c[5];
int m;
}node[N];
int n,m;
int lowbit(int x) {return x&-x;}
int getsum(int x) {
int s=0;
for(;x>0;s+=cal[x],x-=lowbit(x));
return s;
}
void update(int x,int value) {
for(;x<m;cal[x]+=value,x+=lowbit(x));
}
//数组中一共有m个数
int seach(int value) {
int l=1,r=m-1,mid;
while(l<=r)
{
mid=(l+r)>>1;
int t=ptr[mid];
if(t==value) return mid;
else if(t>value)
r=mid-1;
else l=mid+1;
}
return -1;
}
int find(int k) {
int cnt=0,sum=0;
for (int i=20;i>=0;i--) {
sum+=(1<<i);
if(sum>=m||cnt+cal[sum]>=k)
sum-=(1<<i);
else cnt+=cal[sum];
}
return sum+1;
}
int main() {
int i,j,k,t;
char le[5];
while(~scanf("%d%d",&n,&k))
{
memset(cal,0,sizeof(cal));
for(i=1,m=1;i<=n;i++)
{
scanf("%s",node[i].c);
if(node[i].c[0]=='I')
{
scanf("%d",&node[i].m);
ptr[m++]=node[i].m;
}
}
//相当于离散化处理
sort(ptr+1,ptr+m);
//去重
for(i=2,j=1;i<m;i++)
{
if(ptr[i]!=ptr[j])
ptr[++j]=ptr[i];
}
m=j+1;
for(i=1,j=0;i<=n;i++)
{
if(node[i].c[0]=='I')
{
t=seach(node[i].m);
update(t,1); j++;
}
else
{
t=find(j+1-k);
printf("%d\n",ptr[t]);
}
}
}
return 0;
}
附上别人其他做法（来自：http://blog.sina.com.cn/s/blog_732dd9320100trmj.html）：
SBT解法：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<vector>
using namespace std;
const int N=1000005;
struct TNode {
int left,right,size,key;
void init() {
left=0; right=0; size=1;
}
}node[N];
int n,m,tot,root;
void leftrorate(int &t) {//左旋
int k=node[t].right;
node[t].right=node[k].left;
node[k].left=t;
node[k].size=node[t].size;
node[t].size=node[node[t].left].size+node[node[t].right].size+1;
t=k;
}
void rightrorate(int &t) {//右旋
int k=node[t].left;
node[t].left=node[k].right;
node[k].right=t;
node[k].size=node[t].size;
node[t].size=node[node[t].left].size+node[node[t].right].size+1;
t=k;
}
void maintain(int &t,bool flag) {//维护
if(!flag) {
if(node[node[node[t].left].left].size>node[node[t].right].size)
rightrorate(t);
else if(node[node[node[t].left].right].size>node[node[t].right].size) {
leftrorate(node[t].left);
rightrorate(t);
}
else return ;
}
else {
if(node[node[node[t].right].right].size>node[node[t].left].size)
leftrorate(t);
else if(node[node[node[t].right].left].size>node[node[t].left].size) {
rightrorate(node[t].right);
leftrorate(t);
}
else return ;
}
maintain(node[t].left,0);
maintain(node[t].right,1);
maintain(t,0);
maintain(t,1);
}
void insert(int &t,int value) {//插入
if(!t) {
t=++tot;
node[t].init();
node[t].key=value;
}
else {
node[t].size++;
if(value<node[t].key) insert(node[t].left,value);
else insert(node[t].right,value);
maintain(t,value>=node[t].key);
}
}
int find(int t,int k) {//查找第k小的数
if(k<=node[node[t].left].size) return find(node[t].left,k);
else if(k>node[node[t].left].size+1)
return find(node[t].right,k-node[node[t].left].size-1);
else return node[t].key;
}
int main() {
int k,end;
char le[5];
while(~scanf("%d%d",&n,&end)) {
root=tot=0;
for(k=0;k<n;k++) {
scanf("%s",le);
switch(le[0]) {
case 'I':scanf("%d",&m);
insert(root,m); break;
default :
printf("%d\n",find(root,tot+1-end));
}
}
}
return 0;
}
树状数组解法：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;
const int N=1000005;
int cal[N],ptr[N];
struct TNode {
char c[5];
int m;
}node[N];
int n,m;
int lowbit(int x) {return x&-x;}
int getsum(int x) {
int s=0;
for(;x>0;s+=cal[x],x-=lowbit(x));
return s;
}
void update(int x,int value) {
for(;x<m;cal[x]+=value,x+=lowbit(x));
}
int seach(int value) {
int l=1,r=m-1,mid;
while(l<=r) {
mid=(l+r)>>1;
int t=ptr[mid];
if(t==value) return mid;
else if(t>value) r=mid-1;
else l=mid+1;
}
return -1;
}
int find(int k) {
int cnt=0,sum=0;
for (int i=20;i>=0;i--) {
sum+=(1<<i);
if(sum>=m||cnt+cal[sum]>=k)
sum-=(1<<i);
else cnt+=cal[sum];
}
return sum+1;
}
int main() {
int i,j,k,t;
char le[5];
while(~scanf("%d%d",&n,&k)) {
memset(cal,0,sizeof(cal));
for(i=1,m=1;i<=n;i++) {
scanf("%s",node[i].c);
if(node[i].c[0]=='I') {
scanf("%d",&node[i].m);
ptr[m++]=node[i].m;
}
}
sort(ptr+1,ptr+m);
for(i=2,j=1;i<m;i++) {
if(ptr[i]!=ptr[j]) ptr[++j]=ptr[i];
}
m=j+1;
for(i=1,j=0;i<=n;i++) {
if(node[i].c[0]=='I') {
t=seach(node[i].m);
update(t,1); j++;
}
else {
t=find(j+1-k);
printf("%d\n",ptr[t]);
}
}
}
return 0;
}