Educational Codeforces Round 71 (Rated for Div. 2) D

原文链接：https://www.cnblogs.com/xwl3109377858/p/11405773.html

Educational Codeforces Round 71 (Rated for Div. 2)

D - Number Of Permutations

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:

1. if p=[1,2,3], then s=[(1,1),(2,2),(3,1)] — bad sequence (sorted by first elements);
2. if p=[1,3,2], then s=[(1,1),(3,1),(2,2)] — bad sequence (sorted by second elements);
3. if p=[2,1,3], then s=[(2,2),(1,1),(3,1)] — good sequence;
4. if p=[2,3,1], then s=[(2,2),(3,1),(1,1)] — good sequence;
5. if p=[3,1,2], then s=[(3,1),(1,1),(2,2)] — bad sequence (sorted by second elements);
6. 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] [1,1] 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。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<list>
//#include<unordered_map>
using namespace std;
#define ll long long
const int inf=1e9+7;
 
const int maxn=3e5+10;
const int mod=998244353;
ll int jiecheng[maxn];//记录阶乘 

pair<int,int> p[maxn];//pair数组 

map<pair<int,int>,int>mp1;//记录重复二元组数 

map<int,int>x;//标记 
map<int,int>y;

bool cmp(const pair<int,int> &a,const pair<int,int> &b)
{                        //令x递增顺序排 
    if(a.first != b.first)
        return a.first < b.first;
    else
        return a.second < b.second;
}

int main()
{
    ios::sync_with_stdio(false);
    
    int n;
    cin>>n;
    
    x.clear();//初始化 
    y.clear();
    mp1.clear();
    
    jiecheng[0]=1;
    for(int i=1;i<=n;i++)
    {
        cin>>p[i].first>>p[i].second;
        jiecheng[i]=jiecheng[i-1]*i%mod;//算阶乘 
        x[p[i].first]++;//标记 
        y[p[i].second]++;
        mp1[p[i]]++;//标记二元组 
    }
    
    sort(p+1,p+n+1,cmp);//排序 
    
    ll int flag=1;
    
    for(int i=2;i<=n;i++)//看x，y是否同时递增 
    {
        if(!(p[i-1].first<=p[i].first&&p[i-1].second<=p[i].second))
        {
            flag=0;//不符合 
            break;
        }
    }
    ll int ans=0;
    
    if(flag==1)//x,y同时递增 
    {
        for(auto it=mp1.begin();it!=mp1.end();it++)
            flag=flag*jiecheng[it->second]%mod;
        ans=mod-flag;//减去重复情况 
    }
    
    ll int res=1;
    
    for(auto it=x.begin();it!=x.end();it++)//累加x 
        res=res*jiecheng[it->second]%mod;
    ans=(ans+res)%mod;
    
    res=1;
    for(auto it=y.begin();it!=y.end();it++)//累加y 
        res=res*jiecheng[it->second]%mod;
    ans=(ans+res)%mod;
    
    ans=(jiecheng[n]-ans+mod)%mod;//全排列减去所有递增为最终答案 
    
    cout<<ans<<endl;
    
    return 0;
}