A - 4302. 元素分类
Code
cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ll n, x, sum = 0;
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> x;
sum += abs(x);
}
cout << sum;
return 0;
}B - 4303. 链表
Code
cpp
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
int n;
unordered_map<string,string> m;
string s1, s2;
int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
bool flag = false;
cin >> s1 >> s2;
for (auto &item : m)
if (item.y == s1)
{
item.y = s2;
flag = true;
break;
}
if (!flag) m[s1] = s2;
}
cout << m.size() << endl;
for (auto item : m)
cout << item.x << ' ' << item.y << endl;
return 0;
}C - 4304. 字符串归类(并查集)
题目描述

解题思路
对于每个字符串中的字符用id数组来比较 如果之前没出现过,就赋值为i(第几个字符串) 如果出现过,就合并。(合并之后仍然要遍历这个字符串,目的是为了完善id数组)
Code
cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200020;
int n, p[N], id[30];
string str;
int find(int x)
{
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
int main()
{
cin >> n;
for (int i = 1; i <= n; i++) p[i] = i;
int res = n;
for (int i = 1; i <= n; i++)
{
cin >> str;
for (auto x : str)
{
if (id[x - 'a'])
{
if (find(i) != find(id[x - 'a']))
{
p[find(i)] = find(id[x - 'a']);
res--;
}
}
else id[x - 'a'] = i;
}
}
cout << res;
return 0;
}
