1097. Deduplication on a Linked List (25)

//思路：第一步用node将链表链接起来 存在v中 用ma判断是否重复 重复则pop 并push到re中

#include <iostream>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
using namespace std;

struct node {
    int adress, value, next;
} nodes[100001];

int main(){
    int s, n;
    vector<node> v, re;
    map<int, int> ma;

    cin >> s >> n;
    for (int i = 0; i < n; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        nodes[a] = {a, b, c};
    }

    int head = s;
    while (head != -1) {
        v.push_back(nodes[head]);
        int t = abs(nodes[head].value);
        if (ma[t]) {
            v.pop_back();
            re.push_back(nodes[head]);
        }else{
            ma[t] = 1;
        }
        head = nodes[head].next;
    }


    if (v.size() > 0) {
        for (int i = 0; i < v.size(); i++) {
            if(i == 0) printf("%05d %d", v[i].adress, v[i].value);
            else printf(" %05d\n%05d %d", v[i].adress, v[i].adress, v[i].value);
        }
        cout << " -1\n";
    }
    if (re.size() > 0) {
        for (int i = 0; i < re.size(); i++) {
            if(i == 0) printf("%05d %d", re[i].adress, re[i].value);
            else printf(" %05d\n%05d %d", re[i].adress, re[i].adress, re[i].value);
        }
        cout << " -1\n";
    }

    return 0;
}