HDOJ--4006--The kth great number

题目描述：
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.
输入描述:
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.
输出描述:
The output consists of one integer representing the largest number of islands that all lie on one line.
输入：
8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q
输出：
1
2
3
题意：
给定一系列的数，并能够不断往里添加数使这个序列得到更新，问第k大的数是几
题解：
1.运用优先队列将其由小到大保存，令队列里的数据始终只有k个，那么队首元素就是最小值

2.运用堆排列，同样也只是在堆中存放k个元素，将最小值的位置放在堆顶，当超过k个的元素加入时，如果加入的数大于堆顶位置对应的元素，那么将那个数位置上的数更新掉并重排堆，如果小于就不管它了
代码：

#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int main(){
    int n,k,x;
    char c;
    while(scanf("%d%d",&n,&k)!=EOF){
        priority_queue<int,vector<int>,greater<int> > q;
        for(int i=0;i<n;i++){
            cin>>c;
            if(c=='I'){
                scanf("%d",&x);
                q.push(x);
                int len=q.size();
                if(len>k) q.pop();
            }
            else printf("%d\n",q.top());
        }
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<set>
using namespace std;

multiset<int>s;

int main()
{
    int n,k;
    char str[5];
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        s.clear();
        while(n--)
        {
            scanf("%s",str);
            if(str[0]=='I')
            {
                int w;
                scanf("%d",&w);
                s.insert(w);
                multiset<int>::iterator it;
                while(s.size()>k)
                {
                    it=s.begin();
                    s.erase(it);
                }
            }
            else
            {
                multiset<int>::iterator it;
                 it=s.begin();
                 printf("%d\n",*it);
            }
        }
    }
    return 0;
}

来源：CSDN

作者：ypopstar

链接：https://blog.csdn.net/Ypopstar/article/details/104813308

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