Codeforces 1152E Neko and Flashback 欧拉路径

Neko and Flashback

把a[ i ] - b[ i ] 看成边， 就是求一遍欧拉路径就好了。

注意图不连通的情况。。

#include<bits/stdc++.h>
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
using namespace std;

const int N = (int)2e5 + 7;

int n, a[N], b[N], deg[N];
int hs[N], hs_cnt;
bool ban[N];
vector<PII> G[N];
vector<int> ans;

void dfs(int u) {
    while(G[u].size()) {
        int v = G[u].back().se;
        int id = G[u].back().fi;
        G[u].pop_back();
        if(!ban[id]) {
            ban[id] = true;
            dfs(v);
        }
    }
    ans.push_back(hs[u]);
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        scanf("%d", &a[i]);
        hs[++hs_cnt] = a[i];
    }
    for(int i = 1; i < n; i++) {
        scanf("%d", &b[i]);
        hs[++hs_cnt] = b[i];
    }
    sort(hs + 1, hs + 1 + hs_cnt);
    hs_cnt = unique(hs + 1, hs + 1 + hs_cnt) - hs - 1;
    for(int i = 1; i < n; i++) {
        a[i] = lower_bound(hs + 1, hs + 1 + hs_cnt, a[i]) - hs;
        b[i] = lower_bound(hs + 1, hs + 1 + hs_cnt, b[i]) - hs;
        deg[a[i]]++;
        deg[b[i]]++;
        G[a[i]].push_back(mk(i, b[i]));
        G[b[i]].push_back(mk(i, a[i]));
    }
    for(int i = 1; i < n; i++) {
        if(a[i] > b[i]) {
            puts("-1");
            return 0;
        }
    }
    int be = 1, cnt = 0;
    for(int i = 1; i <= hs_cnt; i++) {
        if(deg[i] & 1) {
            be = i;
            cnt++;
        }
    }
    if(cnt != 0 && cnt != 2) puts("-1");
    else {
        dfs(be);
        if(ans.size() != n) return puts("-1"), 0;
        for(int i = (int)ans.size() - 1; i >= 0; i--) {
            printf("%d%c", ans[i], " \n"[i == 0]);
        }
    }
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/11615194.html