一道vector的题,超时

zhaoqianwan / 2023-05-14 / 原文

“单身狗”是中文对于单身人士的一种爱称。本题请你从上万人的大型派对中找出落单的客人，以便给予特殊关爱。

输入格式：

输入第一行给出一个正整数 N（

输出格式：

首先第一行输出落单客人的总人数；随后第二行按 ID 递增顺序列出落单的客人。ID 间用 1 个空格分隔，行的首尾不得有多余空格。

输入样例：

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

输出样例：

5
10000 23333 44444 55555 88888

代码长度限制

16 KB

时间限制

200 ms

内存限制

64 MB

#include<iostream>

#include<algorithm>

#include<vector>

using·namespace·std;

int·main()

{

····int·N,·b,·c,·M,·e;

····cin·>>·N;

····vector<int>V1;

····vector<int>V2;

····vector<int>V3;

····for·(int·i·=·0;·i·<·N;·i++)

····{

········cin·>>·b·>>·c;

········V1.push_back(b);

········V2.push_back(c);

····}

········cin·>>·M;

····for·(int·i·=·0;·i·<·M;·i++)

····{

········cin·>>·e;

········V3.push_back(e);

····}

····for·(int·i·=·0;·i·<·N;·i++)

····{

········int·count·=·0;

········for·(int·j·=·0;·j·<·V3.size();·j++)

········{

············

············if·(V1[i]·==·V3[j]||V2[i]·==·V3[j])

············{

················count++;

············}

············if·(count·==·2)

············{

················V3.erase(V3.begin()·+·j);//不要用迭代器容易错

················V3.erase(V3.begin()·+·j-1);//因为元素少了一个

················break;

············}

········}

····}

····sort(V3.begin(),·V3.end());

····cout·<<·V3.size()·<<·endl;

····for·(int·i·=·0;·i·<·V3.size();·i++)

····{

········if·(i·!=·V3.size()·-·1)

········{

············cout·<<·V3[i]·<<·"·";

········}

········else

········{

············cout·<<·V3[i];

········}

····}

····return·0;

}