priority_queue：优先队列-白红宇

priority_queue：优先队列

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发布时间：2019-03-04

本文共 2552 字，大约阅读时间需要 8 分钟。

#include

模板声明带3个参数：priority_queue<Type, Container, Functional>，其中Type 为数据类型，Container为保存数据的容器，Functional 为元素比较方式。

Container必须是用数组实现的容器，比如vector,qeque等等，但不能用 list。STL里面默认用的是vector。

比较方式默认用operator<，所以如果把后面2个参数缺省的话，优先队列就是降序，队头元素最大。

第三个参数是一个比较类，less 表示数字大的优先级高，而 greater 表示数字小的优先级高。

尽管队列和优先队列都是在queue里的，但是队列的队首是front来获得，而优先队列的队首是top来获得

HDU 4006 The kth great number

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

I 5

I 4

Sample Output

Hint

Xiao 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).

#include 
   
    #include 
    
     #include 
     
       #include 
      
       #include 
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            using namespace std;int main(){ int n, k, t; char s[5]; while (scanf("%d%d", &n, &k)!=EOF){ priority_queue

, greater

> q; for (int i=1; i<=n; i++){ scanf("%s", s); if (s[0]=='I'){ scanf("%d", &t); q.push(t); if (q.size()>k) q.pop(); }else printf("%d\n", q.top()); } } return 0;}

HDU 3785 寻找大富翁

浙江桐乡乌镇共有n个人,请找出该镇上的前m个大富翁.

Input

输入包含多组测试用例.

每个用例首先包含2个整数n（0<n<=100000）和m(0<m<=10)，其中: n为镇上的人数，m为需要找出的大富翁数, 接下来一行输入镇上n个人的财富值.

n和m同时为0时表示输入结束.

Output

请输出乌镇前m个大富翁的财产数，财产多的排前面，如果大富翁不足m个，则全部输出,每组输出占一行.

Sample Input

3 1

2 5 -1

5 3

1 2 3 4 5

0 0

Sample Output

5 4 3

#include 
   
    #include 
    
     #include 
     
       #include 
      
       #include 
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            using namespace std;int main(){ int n, m, t; while (scanf("%d%d", &n, &m)!=EOF){ if (n==0 && m==0) break; priority_queue

, less

> q; for (int i=1; i<=n; i++){ scanf("%d", &t); q.push(t); } if (q.size()>m){ printf("%d", q.top()); q.pop(); for (int i=2; i<=m; i++){ printf(" %d", q.top()); q.pop(); } }else{ printf("%d", q.top()); while (!q.empty()){ printf(" %d", q.top()); } } printf("\n"); } return 0;}

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