原文链接：https://www.cnblogs.com/xwl3109377858/p/11405773.html
Educational Codeforces Round 71 (Rated for Div. 2)
You are given a sequence of n pairs of integers: (a1,b1),(a2,b2),…,(an,bn). This sequence is called bad if it is sorted in non-descending order by first elements or if it is sorted in non-descending order by second elements. Otherwise the sequence is good. There are examples of good and bad sequences:
- s=[(1,2),(3,2),(3,1)] is bad because the sequence of first elements is sorted: [1,3,3];
- s=[(1,2),(3,2),(1,2)] is bad because the sequence of second elements is sorted: [2,2,2];
- s=[(1,1),(2,2),(3,3)] is bad because both sequences (the sequence of first elements and the sequence of second elements) are sorted;
- s=[(1,3),(3,3),(2,2)] is good because neither the sequence of first elements ([1,3,2]) nor the sequence of second elements ([3,3,2]) is sorted.
Calculate the number of permutations of size n such that after applying this permutation to the sequence s it turns into a good sequence.
A permutation p of size n is a sequence p1,p2,…,pn consisting of n distinct integers from 1 to n (1≤pi≤n). If you apply permutation p1,p2,…,pn to the sequence s1,s2,…,sn you get the sequence sp1,sp2,…,spn. For example, if s=[(1,2),(1,3),(2,3)] and p=[2,3,1] then s turns into [(1,3),(2,3),(1,2)].
Input
The first line contains one integer n (1≤n≤3⋅105).
The next n lines contains description of sequence s. The i-th line contains two integers ai and bi (1≤ai,bi≤n) — the first and second elements of i-th pair in the sequence.
The sequence s may contain equal elements.
Output
Print the number of permutations of size n such that after applying this permutation to the sequence s it turns into a good sequence. Print the answer modulo 998244353 (a prime number).
Examples
input
3
1 1
2 2
3 1
output
3
input
4
2 3
2 2
2 1
2 4
output
0
input
3
1 1
1 1
2 3
output
4
Note
In first test case there are six permutations of size 3:
- if p=[1,2,3], then s=[(1,1),(2,2),(3,1)] — bad sequence (sorted by first elements);
- if p=[1,3,2], then s=[(1,1),(3,1),(2,2)] — bad sequence (sorted by second elements);
- if p=[2,1,3], then s=[(2,2),(1,1),(3,1)] — good sequence;
- if p=[2,3,1], then s=[(2,2),(3,1),(1,1)] — good sequence;
- if p=[3,1,2], then s=[(3,1),(1,1),(2,2)] — bad sequence (sorted by second elements);
- if p=[3,2,1], then s=[(3,1),(2,2),(1,1)] — good sequence.
题意：题目要求使二元组序列中x和y都不递增的排列的数量，
例如二元组 [1,1] [2,2] [3,1] 有三种 [2,2] [1,1] [3,1]
[2,2] [3,1] [1,1] [3,1] [2,2] [3,1] 符合条件。
思路：考虑到求不递增的排列比较难，我们可以反过来考虑求递增的排列，
然后用全排列的数量即阶乘减去递增序列数量即为答案。
求递增的排列时要考虑单个x和y排列中重复元素数量，
而x与y同时递增时有重复的二元组我们应该减掉。
例如 [1,2] [3,4] [3,4] [5,6]，计算单个排列x和y里计算后排列数量都为2，
但是二元组重复了排列数量2，所以答案为n阶乘减去2而不是减4。
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<map>
7 #include<set>
8 #include<vector>
9 #include<queue>
10 #include<stack>
11 #include<list>
12 //#include<unordered_map>
13 using namespace std;
14 #define ll long long
15 const int inf=1e9+7;
16
17 const int maxn=3e5+10;
18 const int mod=998244353;
19 ll int jiecheng[maxn];//记录阶乘
20
21 pair<int,int> p[maxn];//pair数组
22
23 map<pair<int,int>,int>mp1;//记录重复二元组数
24
25 map<int,int>x;//标记
26 map<int,int>y;
27
28 bool cmp(const pair<int,int> &a,const pair<int,int> &b)
29 { //令x递增顺序排
30 if(a.first != b.first)
31 return a.first < b.first;
32 else
33 return a.second < b.second;
34 }
35
36 int main()
37 {
38 ios::sync_with_stdio(false);
39
40 int n;
41 cin>>n;
42
43 x.clear();//初始化
44 y.clear();
45 mp1.clear();
46
47 jiecheng[0]=1;
48 for(int i=1;i<=n;i++)
49 {
50 cin>>p[i].first>>p[i].second;
51 jiecheng[i]=jiecheng[i-1]*i%mod;//算阶乘
52 x[p[i].first]++;//标记
53 y[p[i].second]++;
54 mp1[p[i]]++;//标记二元组
55 }
56
57 sort(p+1,p+n+1,cmp);//排序
58
59 ll int flag=1;
60
61 for(int i=2;i<=n;i++)//看x，y是否同时递增
62 {
63 if(!(p[i-1].first<=p[i].first&&p[i-1].second<=p[i].second))
64 {
65 flag=0;//不符合
66 break;
67 }
68 }
69 ll int ans=0;
70
71 if(flag==1)//x,y同时递增
72 {
73 for(auto it=mp1.begin();it!=mp1.end();it++)
74 flag=flag*jiecheng[it->second]%mod;
75 ans=mod-flag;//减去重复情况
76 }
77
78 ll int res=1;
79
80 for(auto it=x.begin();it!=x.end();it++)//累加x
81 res=res*jiecheng[it->second]%mod;
82 ans=(ans+res)%mod;
83
84 res=1;
85 for(auto it=y.begin();it!=y.end();it++)//累加y
86 res=res*jiecheng[it->second]%mod;
87 ans=(ans+res)%mod;
88
89 ans=(jiecheng[n]-ans+mod)%mod;//全排列减去所有递增为最终答案
90
91 cout<<ans<<endl;
92
93 return 0;
94 }