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;}