mp

问题 G: Green Bin

时间限制: 1 Sec  内存限制: 128 MB
[提交] [状态]

题目描述

We will call a string obtained by arranging the characters contained in a string a in some order, an anagram of a.
For example, greenbin is an anagram of beginner. As seen here, when the same character occurs multiple times, that character must be used that number of times.
Given are N strings s1,s2,…,sN. Each of these strings has a length of 10 and consists of lowercase English characters. Additionally, all of these strings are distinct. Find the number of pairs of integers i,j (1≤i<j≤N) such that si is an anagram of sj.

Constraints
·2≤N≤10⁵
·si is a string of length 10.
·Each character in si is a lowercase English letter.
·s1,s2,…,sN are all distinct.

输入

Input is given from Standard Input in the following format:

N
s1
s2
:
sN

输出

Print the number of pairs of integers i,j (1≤i<j≤N) such that si is an anagram of sj.

样例输入
Copy

【样例1】
3
acornistnt
peanutbomb
constraint
【样例2】
2
oneplustwo
ninemodsix
【样例3】
5
abaaaaaaaa
oneplustwo
aaaaaaaaba
twoplusone
aaaabaaaaa

样例输出 Copy

【样例1】
1
【样例2】
0
【样例3】
4

提示

样例1解释
s1= acornistnt is an anagram of s3= constraint. There are no other pairs i,j such that si is an anagram of sj, so the answer is 1.
样例2解释
If there is no pair i,j such that si is an anagram of sj, print 0.
样例3解释
Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

解析：

题意概括：

找出有这些单词所含的字母种类和数目都相等的对数。注意单词之间可以随意组合，因此要用到排列的知识。

解题思路：

首先将字符串内的字符进行从小到大排列，这样就很容易比较两个字符串是否满足题意要求。使用map容器记录相同单词数，最后使用排列知识算出总个数。给你一些字符串，判断其中有多少对字符串调换字符位置后会得到一样的字符串（组成字母及数目一样）
由组合数知识知n个相同组成的字符串可以构成C2n对，即n*（n-1）/2

如果是这样定义

mp[99] = 1000000;

it->first就是下标，即99
it->second是值，即1000000

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
const int maxn=1e6;
string a;
map<string,ll>mp;
int n;
void inint(){
    cin>>n;
    for(int i=;i<n;i++){
        cin>>a;
        sort(a.begin(),a.end());
        mp[a]++;
    }
}
int main(){
    inint();
    ll sum=;
    for(auto i=mp.begin();i!=mp.end();i++){
        ll s=i->second;
        sum+=(s*(s-))/;
    }
    printf("%lld",sum);
}

mp的解析：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
const int maxn=1e6;
string a;
map<string,ll>mp;
int main(){
    cin>>a;
    mp[a]=;
    auto i=mp.begin();
    cout<<i->first<<endl;//输入的字符串
    cout<<i->second;//下标
}

输出：