HDOJ 4006 The kth great number（优先队列）

题目：

The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 3770 Accepted Submission(s): 1581



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
The output consists of one integer representing the largest number of islands that all lie on one line.

[align=left]Sample Input[/align]

8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q

[align=left]Sample Output[/align]

1
2
3

HintXiao  Ming  won't  ask  Xiao  Bao  the  kth  great  number  when  the  number  of  the  written number is smaller than k. (1=<k<=n<=1000000).

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4006

题意：

I表示输入数据，当输入Q的时候就在输入的那些数据中找第K大的数

思路：

刚开始用sort排序，因为数据量大，会超时，所以必须用最小优先队列（降序排序），先把k个数压进队列，之后的数和队头比较，如果比队头校就不进队，若比队头大，就队头出队，然后压进队列里。。。

代码：

/*
sort不行 // 数据大  会超市
队列  最小优先队列

#include<iostream>
#include <algorithm>
using namespace std;
int used[10000+10];
bool cmp(const int a,const int b)
{
return a > b;
}
int main()
{
int n,k;
char ch;
int i=0;
int a[10000+10];
int flag = 0;

while(cin>>n>>k;)
{
memset(used, 0 ,sizeof(used));
cin>>ch;
if(ch == 'I')
{
cin>>a[i++];
}
else if(ch == 'Q')
{
sort(a,a+n,cmp);
cout<<a[k-1-flag]<<endl;
flag++;
}
}

}
*/

#include<iostream>
#include<queue>
using namespace std;

const int maxn = 1000000 + 10;

struct node
{
int t;
const bool operator < (const struct node &a) const
{
return t > a.t;
}
}a[maxn];

priority_queue<struct node > q;

int main()
{
int temp;
int n,k;
char ch;
int i=0;
int flag = 0;
while(cin>>n>>k)
{
while(k--)
{
n--;
cin>>ch;
if(ch == 'I')
{
cin>>a[++i].t;  //不能用a[i++];
q.push(a[i]);
//	i++;
}
}

while(n--)
{
cin>>ch;
if(ch == 'I')
{
cin>>a[++i].t;
temp = q.top().t;
if(a[i].t > temp)
{
q.pop();
q.push(a[i]);
}
//	i++;
}
else if(ch == 'Q')
{
cout<<q.top().t<<endl;
}

}

while(!q.empty())
{
q.pop();
}
}
}