codeforces1207D - Number Of Permutations

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D. Number Of Permutations
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a sequence of ? pairs of integers: (?1,?1),(?2,?2),…,(??,??). 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:

?=[(1,2),(3,2),(3,1)] is bad because the sequence of first elements is sorted: [1,3,3];
?=[(1,2),(3,2),(1,2)] is bad because the sequence of second elements is sorted: [2,2,2];
?=[(1,1),(2,2),(3,3)] is bad because both sequences (the sequence of first elements and the sequence of second elements) are sorted;
?=[(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 ? such that after applying this permutation to the sequence ? it turns into a good sequence.

A permutation ? of size ? is a sequence ?1,?2,…,?? consisting of ? distinct integers from 1 to ? (1≤??≤?). If you apply permutation ?1,?2,…,?? to the sequence ?1,?2,…,?? you get the sequence ??1,??2,…,???. For example, if ?=[(1,2),(1,3),(2,3)] and ?=[2,3,1] then ? turns into [(1,3),(2,3),(1,2)].

Input
The first line contains one integer ? (1≤?≤3⋅105).

The next ? lines contains description of sequence ?. The ?-th line contains two integers ?? and ?? (1≤??,??≤?) — the first and second elements of ?-th pair in the sequence.

The sequence ? may contain equal elements.

Output
Print the number of permutations of size ? such that after applying this permutation to the sequence ? it turns into a good sequence. Print the answer modulo 998244353 (a prime number).

Examples
inputCopy
3
1 1
2 2
3 1
outputCopy
3
inputCopy
4
2 3
2 2
2 1
2 4
outputCopy
0
inputCopy
3
1 1
1 1
2 3
outputCopy
4
Note
In first test case there are six permutations of size 3:

if ?=[1,2,3], then ?=[(1,1),(2,2),(3,1)] — bad sequence (sorted by first elements);
if ?=[1,3,2], then ?=[(1,1),(3,1),(2,2)] — bad sequence (sorted by second elements);
if ?=[2,1,3], then ?=[(2,2),(1,1),(3,1)] — good sequence;
if ?=[2,3,1], then ?=[(2,2),(3,1),(1,1)] — good sequence;
if ?=[3,1,2], then ?=[(3,1),(1,1),(2,2)] — bad sequence (sorted by second elements);
if ?=[3,2,1], then ?=[(3,1),(2,2),(1,1)] — good sequence.

题意： 一些二元组(x,y)。求多少种排列，使得x不递增（包括相等），y不递增（包括相等）。

思路： 比赛的时候一秒就想出了思路，数学能力不行，实现的时候一直出问题。。。

明显的组合问题！！

要考虑多少个不递增的排列是很困难的，正难则反，我们考虑多少个递增的排列，再用全排列减去这个就可以了。
1.先考虑问题的简化版本：如果是一元组x，并且没有相同的情况，怎么求？此时x的递增情况只有一种，其他情况全是非递增的，只需要n的阶乘-1就可以了。
2.再考虑：如果一元组x，有重复的数字，怎么办？此时x的递增情况，与其中的重复数组有关，比如1 2 2。第二个2和第三个2可以调换位置。此时的递增数目为所有重复数目阶乘和积。再用fac[n]减去重复部分就可以了。
3.继续考虑：如果变成了二元组(x,y)，没有重复数字：只考虑x的递增情况，有一种，只考虑y的递增情况，有一种。但是！这是二元组，你算x的时候，忽略了y，算y的时候，忽略了x。其中会有重复的部分。假设（x,y）x,y同时递增。那么x递增的情况和y递增的情况是等价的，要减去1.
4.最后是题目的情况：二元组（x,y），重复数字。只考虑x,y，就是第二种情况的和：重复数目阶乘的积。在考虑第三种情况，重复部分。假设x,y能同时递增，并且出现了相同的二元组，那么这个相同的二元组被计算了两遍，就必须减去相同二元组的情况。

比如（1，2）（2，3）（2，3）（4，5）。
只考虑x是1，2，2，4。有两个2答案是2。
再考虑y是2，3，3，5。有两个3答案是3.
但是你算x的2和你算y的3其实是一种情况——交换2的时候也在交换3，所以要减去二元组的重复部分。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <iostream>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;
const ll MOD = 998244353;
const ll maxn = 4e5 + 7;
ll fac[maxn];
map<ll,ll>vis,vis2;struct Node
{ll x,y;
}nodes[maxn];
pair<ll,ll>N[maxn];
map<pair<ll,ll>,ll>mp;
ll cmp1(Node a,Node b)
{return a.x < b.x;
}ll cmp2(Node a,Node b)
{if(a.x == b.x)return a.y < b.y;return a.x < b.x;
}void init()
{fac[0] = 1;for(ll i = 1;i <= 400000;i++){fac[i] = (fac[i - 1] * i) % MOD;}
}int main()
{init();ll tmp1 = 0,tmp2 = 0,tmp3 = 0;ll n;scanf("%lld",&n);for(ll i = 1;i <= n;i++){scanf("%lld%lld",&nodes[i].x,&nodes[i].y);vis[nodes[i].x]++;vis2[nodes[i].y]++;if(vis[nodes[i].x] == n){printf("0\n");return 0;}else if(vis2[nodes[i].y] == n){printf("0\n");return 0;}}ll ans = 0;ll tmp = 1;for(int i = 1;i < maxn;i++)//感谢 F-幕：大佬的提醒，一开始写了 i<=maxn，其实越界了{if(vis[i])tmp = (tmp * fac[vis[i]] + MOD) % MOD;}ans = (ans + tmp) % MOD;tmp1 = tmp;tmp = 1;for(int i = 1;i < maxn;i++){if(vis2[i]){tmp = (tmp * fac[vis2[i]] + MOD) % MOD;}}tmp2 = tmp;ans = (ans + tmp) % MOD;sort(nodes + 1,nodes + 1 + n,cmp2);int ok = 1;for(int i = 2;i <= n;i++){if(nodes[i].x < nodes[i - 1].x || nodes[i].y < nodes[i - 1].y){ok = 0;break;}}for(int i = 1;i <= n;i++){N[i].first = nodes[i].x;N[i].second = nodes[i].y;}if(ok == 1){tmp = 1;for(int i = 1;i <= n;i++){mp[N[i]]++;}map<pair<ll,ll>,ll>::iterator it;for(it = mp.begin();it != mp.end();it ++){tmp = tmp * fac[it -> second] % MOD;}ans = ((ans + MOD) - tmp) % MOD;tmp3 = tmp;}ans = (fac[n] + MOD - ans) % MOD;printf("%lld\n",ans);return 0;
}