B. New Year and Ascent Sequence----思维/水

A sequence a=[a1,a2,…,al] of length l has an ascent if there exists a pair of indices (i,j) such that 1≤i<j≤l and ai<aj. For example, the sequence [0,2,0,2,0] has an ascent because of the pair (1,4), but the sequence [4,3,3,3,1] doesn’t have an ascent.

Let’s call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0,2,0,2,0] and [4,3,3,3,1] is the sequence [0,2,0,2,0,4,3,3,3,1]. The concatenation of sequences p and q is denoted as p+q.

Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s1,s2,…,sn which may have different lengths.

Gyeonggeun will consider all n2 pairs of sequences sx and sy (1≤x,y≤n), and will check if its concatenation sx+sy has an ascent. Note that he may select the same sequence twice, and the order of selection matters.

Please count the number of pairs (x,y) of sequences s1,s2,…,sn whose concatenation sx+sy contains an ascent.

Input
The first line contains the number n (1≤n≤100000) denoting the number of sequences.

The next n lines contain the number li (1≤li) denoting the length of si, followed by li integers si,1,si,2,…,si,li (0≤si,j≤106) denoting the sequence si.

It is guaranteed that the sum of all li does not exceed 100000.

Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.

Examples
inputCopy
5
1 1
1 1
1 2
1 4
1 3
outputCopy
9
inputCopy
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
outputCopy
7
inputCopy
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
outputCopy
72
Note
For the first example, the following 9 arrays have an ascent: [1,2],[1,2],[1,3],[1,3],[1,4],[1,4],[2,3],[2,4],[3,4]. Arrays with the same contents are counted as their occurences.

题意:给你n个序列,任选两个序列串接，序列p和序列q链接，p+q.使ai<aj (1<=i<j<=n)。问你有多少个连接。

解析:对于每个序列我们开两个数组一个记录序列的最大值，一个记录序列的最小值.
特殊情况就是这个序列本身就在这个性质，那么它的最大值为无穷大，最小值为-1.
我们给最大值排序.遍历数组如果最小值<最大值即满足条件，贡献为:当前最大值的位置到末尾有多少个数累加起来即可。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+10;
vector<int> minn,maxn;
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x,k;
		cin>>k;
		bool  f=false;
		int mx=-0x3f3f3f3f;
		int mn=0x3f3f3f3f;
		for(int j=1;j<=k;j++)
		{
			cin>>x;
			if(x>mn)
			{
				f=true;
			}
			mx=max(mx,x);
			mn=min(mn,x);
		}
		if(f==true) {
			minn.push_back(-1);
			maxn.push_back(N);
		}
		else
		{
			minn.push_back(mn);
			maxn.push_back(mx);
		}
	}
	sort(maxn.begin(),maxn.end());
	ll ans=0;
	for(int i=0;i<minn.size();i++)
	{
		ans+=(maxn.end()-upper_bound(maxn.begin(),maxn.end(),minn[i]));
	}
	cout<<ans<<endl;
 }