天梯赛 L2-002 链表去重

发布于:2025-03-18 ⋅ 阅读:(20) ⋅ 点赞:(0)

使用unordered_map记录每个地址的下一个地址,对于每个节点判断以下,已经出现过了,就放到ans数组中,其他的放到ans1中,注意遇到-1就直接结束。

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
typedef long long ll;
const int N = 510;
const int mod = 998244353;
void solve() {
	string st;
	int n;
	cin>>st>>n;
	unordered_map<string,string> next;
	unordered_map<string,int> value;
	for(int i=1;i<=n;i++){
		string s1,s2;
		int w;
		cin>>s1>>w>>s2;
		next[s1] = s2;
		value[s1] = w;
	}
	// for(auto it : next){
		// cout<<it.first<<" "<<it.second<<endl;
// 		
	// }
	unordered_map<int,int> mp;//标记是否被访问
	vector<pair<string,int>> ans1,ans;
	string now = st;
	while(now != "-1"){
		int w = abs(value[now]);
		if(mp[w] >= 1){
			ans.push_back({now,value[now]});
		}else{
			// cout<<"bj"<<now<<endl;
			mp[w]++;
			ans1.push_back({now,value[now]});	
		}
		now = next[now];
	}
	int len = ans1.size();
	ans1.push_back({"-1",-1});
	for(int i = 0 ; i < len; i++){
		cout<<ans1[i].first<<" "<<ans1[i].second
		<<" "<<ans1[i+1].first<<endl;
	}
	len = ans.size();
	ans.push_back({"-1",-1});
	for(int i = 0 ; i < len; i++){
		cout<<ans[i].first<<" "<<ans[i].second
		<<" "<<ans[i+1].first<<endl;
	}
}
signed main() {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int tt = 1;
    //    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}