hdu2275(multiset)

最新推荐文章于 2020-04-20 16:20:04 发布

晓风残月xj

最新推荐文章于 2020-04-20 16:20:04 发布

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分类专栏： STL 文章标签：数据结构 STL

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本文介绍了一种特殊容器的设计与实现方法，该容器支持两种主要操作：压入(push)整数与弹出(pop)不超过指定值的最大整数。通过使用multiset容器，实现了高效的插入、查找及删除操作。

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Kiki & Little Kiki 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 708 Accepted Submission(s): 235

Problem Description

Kiki is considered as a smart girl in HDU, many boys fall in love with her! Now, kiki will finish her education, and leave school, what a pity! One day, zjt meets a girl, who is like kiki very much, in the campus, and calls her little kiki. Now, little kiki want to design a container, which has two kinds of operation, push operation, and pop operation.
Push M:
Push the integer M into the container.
Pop M:
Find the maximal integer, which is not bigger than M, in the container. And pop it from the container. Specially, for all pop operations, M[i] is no bigger than M[i+1].
Although she is as smart as kiki, she still can't solve this problem! zjt is so busy, he let you to help little kiki to solve the problem. Can you solve the problem?

Input

The input contains one or more data sets. At first line of each input data set is an integer N (1<= N <= 100000) indicate the number of operations. Then N lines follows, each line contains a word (“Push” or “Pop”) and an integer M. The word “Push” means a push operation, while “Pop” means a pop operation. You may assume all the numbers in the input file will be in the range of 32-bit integer.

Output

For each pop operation, you should print the integer satisfy the condition. If there is no integer to pop, please print “No Element!”. Please print a blank line after each data set.

Sample Input

  
   9
Push 10
Push 20
Pop 2
Pop 10
Push 5
Push 2
Pop 10
Pop 11
Pop 19
3
Push 2
Push 5
Pop 2

Sample Output

  
   No Element!
10
5
2
No Element!

2

Source

HDU 8th Programming Contest Site（1）

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lcy

本题涉及插入、查找并删除的操作。必须考虑降低时间复杂度，最为有效的方法是平衡二叉树，但是变成复杂度高，其替代品红黑树、伸展树也不是很容易编程实现的。所以想到set、multiset容器来方便编程，其类似红黑树，插入、查找并删除操作的时间复杂度O(log(N))。用insert（）插入m，用lower_bound（）查找大于等于m的元素，用erase（）删除不大于m的元素。注意小细节。

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

char cmd[10];
multiset<int>MulSet;
multiset<int>::iterator iter;

int main()
{
	int n,i,a;
	while(~scanf("%d",&n))
	{
		MulSet.clear();
		while(n--)
		{
			scanf("%s%d",cmd,&a);
			if(cmd[1]=='u')
			{
				MulSet.insert(a);
			}
			else 
			{
				iter=MulSet.lower_bound(a);
				if(*iter==a)
				{
					printf("%d\n",a);
					MulSet.erase(iter);
				}
				else if(*iter!=a)
				{
					if(iter==MulSet.begin())
					{
						printf("No Element!\n");
					}
					else 
					{
						iter--;
						printf("%d\n",*(iter));
						MulSet.erase(iter);
					}
				}
			}
		}
		printf("\n");
	}
	return 0;
}