---
title: "L2-039 清点代码库"
created: 2025-11-28
tags:
  - 算法
---

# L2-039 清点代码库

## 题目 [L2-039 清点代码库](https://pintia.cn/problem-sets/994805046380707840/exam/problems/type/7?problemSetProblemId=1386335159927652362&page=1)

![[image-04fb67d9.png]]

## 思路分析

map计数

转为vector 排序

## 代码实现

算是投机取巧 直接输出带空格的string 而不是多个int

18/25

```cpp
#include <bits/stdc++.h>

using namespace std;

#define endl '\n'

#define int long long

using ll = long long;

using ull = unsigned long long;

using PII = pair<int, int>;

using Pll = pair<ll, ll>;

int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };

const int inf = 0x3f3f3f3f;

using PSI = pair<string,int>;

signed main() {

	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	// 若输入相同 输出相同 则重复

	// 输出简化为整数

	// n m

	int n,m;

	cin>>n>>m;

	unordered_map<string,int> hx;

	while(n--) {

		string s;

		for(int i=0; i<m; i++) {

			int tmp;

			cin>>tmp;

			s+=" ";

			s+=to_string(tmp);

		}

		hx[s]++;

	}

	cout<<hx.size()<<endl;

	vector<PSI> res(hx.begin(),hx.end());

	sort(res.begin(),res.end(),[](const PSI& a,const PSI& b) {

		if (a.second != b.second) return a.second > b.second;

		return a.first < b.first;

	});

	for(auto v:res) {

		cout<<v.second<<v.first<<endl;

	}

	return 0;

}
```

其实也可以直接用数组为键

```cpp
#include <bits/stdc++.h>

using namespace std;

#define endl '\n'

using ll = long long;

using ull = unsigned long long;

using PII = pair<int, int>;

using Pll = pair<ll, ll>;

int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };

const int inf = 0x3f3f3f3f;

using PIV = pair<int,vector<int>>;

signed main() {

	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

	int n,m;cin>>n>>m;

	map<vector<int>,int> hx;

	while(n--) {

		vector<int> cur(m);

		for(int i=0; i<m; i++) {

			cin>>cur[i];

		}

		hx[cur]++;

	}

	cout<<hx.size()<<endl;

	vector<PIV> res;

	for(auto v:hx){

		res.push_back({v.second,v.first});

	}

	sort(res.begin(),res.end(),[](const PIV& a,const PIV& b){

		if(a.first!=b.first)	return a.first>b.first;

		return a.second<b.second;

	});

	for(auto v:res){

		cout<<v.first;

		for(auto n:v.second)	cout<<" "<<n;

		cout<<endl;

	}

	return 0;

}
```

## 同类题型

## 视频讲解

---
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