Codeforces 673D Bear and Two Paths (贪心构造)

题意

给出n个点，给出a,b,c,d，求能不能用少于m的边构成两条欧拉路
一条从a开始到b一条从c开始到d。

思路

因为m只有上限没有下限，我们贪心的构造就可以了，也就是说除了a,b,c,d其他的点都用相同的路径，这样就能使得路径最少了。
需要注意的是n等于4的时候。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n, m, a, b, c, d;
    cin >> n >> m;
    cin >> a >> b >> c >> d;
    if (m < n - 1 + 2 || n == 4) return puts("-1");
    vector<int> v;
    for (int i = 1; i <= n; i++)
    {
        if (i != a && i != b && i != c && i != d)
            v.push_back(i);
    }
    cout << a << " " << c << " ";
    for (int i = 0; i < v.size(); i++)
        cout << v[i] << " ";
    cout << d << " " << b << endl;
    cout << c << " " << a << " ";
    for (int i = 0; i < v.size(); i++)
        cout << v[i] << " ";
    cout << b << " " << d << endl;
    return 0;
}