Similar Arrays CodeForces - 1090D （构造）

Similar Arrays

题意：

题目要构造两个序列

A序列每个数字都不相同，B序列只有两个相同的数字

给M组关系 i ，j表示 ai > aj 或者 ai < aj

同时B序列有这个和A序列一样的关系 如果 ai > aj 就有 bi > bj

思路：

首先考虑不能构造的情况 ，如果A序列中任意一对数字都存在关系，那么B序列中不可能由两个相同的数字

即 m >= (n - 1) * n / 2

如果 m < (n - 1) * n / 2 ，找两个没有关系的点，A序列赋值为n 和 n - 1 ，  B序列赋值为 n 和 n 

剩下的A、B序列都一样就好了

  1 #include<cstdio>
  2 #include<string.h>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<iostream>
  6 #include<vector>
  7 #include<queue>
  8 #include<set>
  9 #include<map>
 10 #include<cctype>
 11 #define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
 12 #define mem(a,x) memset(a,x,sizeof(a))
 13 #define lson rt<<1,l,mid
 14 #define rson rt<<1|1,mid + 1,r
 15 #define P pair<int,int>
 16 #define ull unsigned long long
 17 using namespace std;
 18 typedef long long ll;
 19 const int maxn = 2e5 + 10;
 20 const ll mod = 998244353;
 21 const int inf = 0x3f3f3f3f;
 22 const long long INF = 0x3f3f3f3f3f3f3f3f;
 23 const double eps = 1e-7;
 24 inline ll read()
 25 {
 26     ll X = 0, w = 0; char ch = 0;
 27     while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
 28     while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
 29     return w ? -X : X;
 30 }
 31 ll n, m , a , b;
 32 vector<int>vec[maxn], v;
 33 ll cnt[maxn], arr1[maxn] , arr2[maxn] , cnt1[maxn];
 34 
 35 
 36 
 37 int main()
 38 {
 39     n = read(), m = read();
 40     for (int i = 1; i <= m; ++i)
 41     {
 42         int u = read(), v = read();
 43         vec[u].push_back(v);
 44         vec[v].push_back(u);
 45     }
 46     if (2 * m >= n * (n - 1))
 47     {
 48         cout << "NO" << endl;
 49         return 0;
 50     }
 51     int anspos;
 52     for (int i = 1; i <= n; ++i)
 53     {
 54         if (vec[i].size() == n - 1) continue;
 55         else
 56         {
 57             anspos = i;
 58             for (int j = 0; j < vec[i].size(); ++j) cnt[vec[i][j]] = 1;
 59             break;
 60         }
 61     }
 62     for (int i = 1; i <= n; ++i)
 63     {
 64         if (cnt[i] == 0 && i != anspos)
 65         {
 66             arr2[anspos] = arr2[i] = n;
 67             arr1[anspos] = n, arr1[i] = n - 1;
 68             break;
 69         }
 70     }
 71     int res = 0;
 72     for (int i = 1; i <= n; ++i)
 73     {
 74         if (arr1[i] == 0) arr1[i] = ++res;
 75     }
 76     res = 0;
 77     for (int i = 1; i <= n; ++i)
 78     {
 79         if (arr2[i] == 0) arr2[i] = ++res;
 80     }
 81     cout << "YES" << endl;
 82     for (int i = 1; i <= n; ++i)
 83     {
 84         cout << arr1[i];
 85         if (i == n) cout << endl;
 86         else cout << " ";
 87     }
 88     for (int i = 1; i <= n; ++i)
 89     {
 90         cout << arr2[i];
 91         if (i == n) cout << endl;
 92         cout << " ";
 93     }
 94 
 95 }
 96 
 97 
 98 /*
 99 4 5
100 11 12 13 14 15
101 0 4 10 15 20
102 2 10 30 32 40
103 20 21 22 23 24
104 */

AC代码