HDU4006(优先队列)

最新推荐文章于 2017-07-18 11:08:44 发布

hansy2015

最新推荐文章于 2017-07-18 11:08:44 发布

阅读量388

点赞数

CC 4.0 BY-SA版权

文章标签： hdu

本文链接：https://blog.youkuaiyun.com/gh6267/article/details/54426431

本文介绍了一个基于优先队列实现的算法，用于解决第36届ACM/ICPC大连赛区在线赛中的一道题目。该题目要求帮助角色XiaoBao找到当前列表中的第k大的数。通过使用优先队列结构，可以高效地维护列表中最大的k个数，并在需要时返回第k大的数值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The kth great number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 10531 Accepted Submission(s): 4222

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

Source
The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

#include <bits/stdc++.h>

using namespace std;

struct num{
    int a;
    friend bool operator < (num x,num y){
        return x.a>y.a;
    }
} Edge;
std::vector<num> v;
int main(){
    int n,k;
    priority_queue<num> que;
    char str[10];
    int gh;
    while(~scanf("%d%d",&n,&k)){
        for(int i=0;i<n;i++){
            scanf("%s",str);
            if(str[0]=='I'){
                scanf("%d",&Edge.a);
                que.push(Edge);
                if(que.size()>k){
                    que.pop();
                }
            }
            else{
                printf("%d\n",(que.top()).a);
            }
        }
        while(!que.empty()){
            que.pop();
        }
    }
    return 0;
}