【HDU】-4006-The kth great number（优先队列，好）

The kth great number

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

Total Submission(s): 9936    Accepted Submission(s): 3960


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

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

 

Sample Output

1
2
3

题意：n个数分拨输入，遇Q输出里面第m大的元素。

题解：始终保持队列里有m个元素即可，如果下一个元素大于最大的，加进来，把最小的弹出。如果不大于，加进了没有意义，不要，继续。

#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
int num;
bool friend operator < (node a,node b)
{
return a.num>b.num; //数字从小往大排序,小先进,大后进 (先进先出)
}
}a;
int main()
{
int n,m;
char c;
int ant;
while(~scanf("%d %d",&n,&m))
{
ant=0;
priority_queue<node> q;
for(int i=1;i<=n;i++)
{
getchar();
scanf("%c",&c);
if(c=='I')
{
int t;
ant++;
scanf("%d",&a.num);
if(ant<=m) //不足m个数直接进去
{
q.push(a);
}
else //大于m时，判断是否大于队列里的最大值
{
t=a.num;
a=q.top();
if(t<a.num) //小于不管，继续
continue;
else
{
q.pop(); //新的数大于顶 —最大的，新的数进去,把最小的弹出
a.num=t;
q.push(a); //始终保持队列里有m个元素，输出时输出最大的(顶元素)即可
}
}

}
else
{
a=q.top();
printf("%d\n",a.num);
}
}
}
return 0;
}