c++ STL map简介

首先头文件，

#include<map>

反正我用

#include<bits/stdc++.h>

然后就是创建一个map：

map<string,int>mymap；//类似于一个字典的映射关系，string和int可以改成其他的类型
mymap是你定义的map的名字，可以自己改
string可以改为pair<int,int>来表示两个数之间存在关系

再就是添加元素啦：

//第一种方法是用类似于数组的方法添加：
mymap[0]=hihihi;
//第二种是用insert插入：
mymap.insert(pair<int,string>(1,"hahaha"));
//其它的方法：http://t.zoukankan.com/wuchunming-p-3780677.html

map排序：

//先定义一个结构体来规定排序方式
struct rule{
	bool operator()(int a,int b){
		return a>b;
	}
};


//再将此排序方式加入到map定义中：
map<char,int,rule>a;

然后就是输出啦：

	for(auto it=mymap.begin();it!=mymap.end();it++)
	cout<<it->first<<" "<<it->second;

 map呢有一个自动根据key值排序的功能，也就是it->first，这个很好用哦

看例题：

问题 E: 8.4.4 硬木种类-2

Red Alder Ash Aspen Basswood Ash Beech Yellow Birch Ash Cherry Cottonwood Ash Cypress Red Elm Gum Hackberry White Oak Hickory Pecan Hard Maple White Oak Soft Maple Red Oak Red Oak White Oak Poplan Sassafras Sycamore Black Walnut Willow

Ash 13.7931 Aspen 3.4483 Basswood 3.4483 Beech 3.4483 Black Walnut 3.4483 Cherry 3.4483 Cottonwood 3.4483 Cypress 3.4483 Gum 3.4483 Hackberry 3.4483 Hard Maple 3.4483 Hickory 3.4483 Pecan 3.4483 Poplan 3.4483 Red Alder 3.4483 Red Elm 3.4483 Red Oak 6.8966 Sassafras 3.4483 Soft Maple 3.4483 Sycamore 3.4483 White Oak 10.3448 Willow 3.4483 Yellow Birch 3.4483

map直接秒杀！

#include<bits/stdc++.h>
using namespace std;
int main(){
    map<string,int>mymap;
    string temp;
    int num=0;
    while(getline(cin,temp))
    {
        mymap[temp]++;
        num++;
         
    }
    for(auto it=mymap.begin();it!=mymap.end();it++)
    {
        cout<<it->first<<" ";
        cout<<fixed<<setprecision(4)<<it->second*1.0/num*100<<endl;
    }
}

例题：

C - FF

Problem Statement

Takahashi runs an SNS "Twidai," which has

N users from user 1 through user

N. In Twidai, users can follow or unfollow other users.

Q operations have been performed since Twidai was launched. The i-th (1≤i≤Q) operation is represented by three integers Ti​, Ai​, and

Bi​, whose meanings are as follows:

• If

Ti​=1: it means that user Ai​ follows user Bi​. If user Ai​ is already following user

• Bi​ at the time of this operation, it does not make any change.
• If
• Ti​=2: it means that user Ai​ unfollows user Bi​. If user Ai​ is not following user
• Bi​ at the time of this operation, it does not make any change.
• If
• Ti​=3: it means that you are asked to determine if users Ai​ and Bi​ are following each other. You need to print Yes if user Ai​ is following user Bi​ and user Bi​ is following user
  • Ai​, and No otherwise.

  When the service was launched, no users were following any users.

  Print the correct answers for all operations such that

  Ti​=3 in ascending order of

  i.

  Constraints
• 2≤N≤109
• 1≤Q≤2×105
• Ti​=1,2,3 (1≤i≤Q)
• 1≤Ai​≤N (1≤i≤Q)
• 1≤Bi​≤N (1≤i≤Q)
• Ai​=Bi​ (1≤i≤Q)
• There exists
• i (1≤i≤Q) such that
  • Ti​=3.
  • All values in the input are integers.

  ---

  Input

  The input is given from Standard Input in the following format:

  N 

  Q
  

  T1​ 

  A1​ 

  B1​
  

  T2​ 

  A2​ 

  B2​
  

  ⋮
  

  TQ​ 

  AQ​ 

  BQ​
  

  Output

  Print

  X lines, where X is the number of i's (1≤i≤Q) such that Ti​=3. The j-th (1≤j≤X) line should contain the answer to the j-th operation such that

  Ti​=3.

  ---

  Sample Input 1

  Copy

  3 9
  1 1 2
  3 1 2
  1 2 1
  3 1 2
  1 2 3
  1 3 2
  3 1 3
  2 1 2
  3 1 2
  

  Sample Output 1

  Copy

  No
  Yes
  No
  No
  

  Twidai has three users. The nine operations are as follows.

  • User
  1 follows user
• 2. No other users are following or followed by any users.
• Determine if users
• 1 and 2 are following each other. User 1 is following user 2, but user 2 is not following user
• 1, so No is the correct answer to this operation.
• User
• 2 follows user
• 1.
• Determine if users
• 1 and 2 are following each other. User 1 is following user 2, and user 2 is following user
• 1, so Yes is the correct answer to this operation.
• User
• 2 follows user
• 3.
• User
• 3 follows user
• 2.
• Determine if users
• 1 and 3 are following each other. User 1 is not following user 3, and user 3 is not following user
• 1, so No is the correct answer to this operation.
• User
• 1 unfollows user
• 2.
• Determine if users
• 1 and 2 are following each other. User 2 is following user 1, but user 1 is not following user
  • 2, so No is the correct answer to this operation.

  ---

  Sample Input 2

  Copy

  2 8
  1 1 2
  1 2 1
  3 1 2
  1 1 2
  1 1 2
  1 1 2
  2 1 2
  3 1 2
  

  Sample Output 2

  Copy

  Yes
  No
  

  A user may follow the same user multiple times.

  ---

  Sample Input 3

  Copy

  10 30
  3 1 6
  3 5 4
  1 6 1
  3 1 7
  3 8 4
  1 1 6
  2 4 3
  1 6 5
  1 5 6
  1 1 8
  1 8 1
  2 3 10
  1 7 6
  3 5 6
  1 6 7
  3 6 7
  1 9 5
  3 8 6
  3 3 8
  2 6 9
  1 7 1
  3 10 8
  2 9 2
  1 10 9
  2 6 10
  2 6 8
  3 1 6
  3 1 8
  2 8 5
  1 9 10
  

  Sample Output 3

  Copy

  No
  No
  No
  No
  Yes
  Yes
  No
  No
  No
  Yes
  Yes
  

   map直接秒杀：
   

  #include<bits/stdc++.h>
  using namespace std;
  
  
  int main()
  {
  	int n,q;
  	cin>>n>>q;
  	map<pair<int,int>,bool>mp;
  	while(q--)
  	{
  		int flag=0;
  		cin>>flag;
  		if(flag==1)
  		{
  			int a,b;
  			cin>>a>>b;
  			mp[{a,b}]=1;
  		}
  		if(flag==2)
  		{
  			int a,b;
  			cin>>a>>b;
  			mp[{a,b}]=0;
  		}
  		if(flag==3)
  		{
  			int a,b;
  			cin>>a>>b;
  			if(mp[{a,b}]&&mp[{b,a}])
  			cout<<"Yes"<<endl;
  			else
  			cout<<"No"<<endl;
  		}
  	}
  }