hdu 4006 The kth great number

The kth great number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 5929    Accepted Submission(s): 2404

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

Hint

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

 

 

解法一：用优先队列

#include"stdio.h"
#include"queue"
using namespace std;
struct node 
{
	int x;
	friend bool operator<(node a,node b)
	{
		return a.x>b.x;
	}
};
int main()
{
	int n,k,i;
	char ch;
	while(scanf("%d%d",&n,&k)!=-1)
	{
		priority_queueq; //注意：因为是多实例，故把队列定义循环里面可以清空队列
    	node p;
		while(n--)
		{
			getchar();
			scanf("%c",&ch);
			if(ch=='I')
			{
				scanf("%d",&i);
				p.x=i;
				q.push(p);
				if(q.size()>k)
					q.pop();
			}
			else
			{
				p=q.top();
				printf("%d\n",p.x);
			}
		}
	}
	return 0;
}

解法二：可以把数组控制大小为k,前k个元素挨个输入数组；

满足k个元素后快排，从大到小，第K个元素就是第K小的；

然后，每输入一个元素和第K小元素比较大小，维护数组；

#include
#include
#include
#define N 1000010
static int a[N];
int n,k,m;
int cmp(const void *a,const void*b)
{
    return *(int *)a>*(int *)b?-1:1;   
}
void inti(int t)
{
    int i,j;
    if(ma[k])
        {
            for(i=k;i>=0;i--)
                if(a[i]>t)
                    break;
            for(j=k;j>i+1;j--)
                a[j]=a[j-1];
            a[i+1]=t;
        }
    }
}
int main()
{
    int t;
    char ch;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        k--;
        getchar();
        m=0;
        while(n--)
        {
            scanf("%c",&ch);
            if(ch=='I')
            {
                scanf("%d",&t);
                inti(t);
            }
            else
                printf("%d\n",a[k]);
            getchar();
        }
    }
    return 0;
}