1097 Deduplication on a Linked List (25 分)【难度: 一般 / 知识点: 链表 模拟】

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https://pintia.cn/problem-sets/994805342720868352/problems/994805369774129152

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
unordered_map<int,int>ne,val,vis;
int st,n;
vector< pair<int,int> >a,b;
int main(void)
{
    
    
    cin>>st>>n;
    for(int i=0;i<n;i++)
    {
    
    
        int a,b,c; cin>>a>>b>>c;
        val[a]=b,ne[a]=c;
    }
    while(st!=-1)
    {
    
    
        if(vis[abs(val[st])]) b.push_back({
    
    st,val[st]});
        else a.push_back({
    
    st,val[st]}),vis[abs(val[st])]=1;
        st=ne[st];
    }
    for(int i=0;i<a.size();i++)
    {
    
    
        if(i==a.size()-1) printf("%05d %d -1\n",a[i].first,a[i].second);
        else printf("%05d %d %05d\n",a[i].first,a[i].second,a[i+1].first);
    }
    for(int i=0;i<b.size();i++)
    {
    
    
        if(i==b.size()-1) printf("%05d %d -1",b[i].first,b[i].second);
        else printf("%05d %d %05d\n",b[i].first,b[i].second,b[i+1].first);
    }
    return 0;
}

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Origin blog.csdn.net/qq_46527915/article/details/121411172