New Year and Ascent Sequence（二分）

A sequence a=[a1,a2,…,al]a=[a1,a2,…,al] of length ll has an ascent if there exists a pair of indices (i,j)(i,j) such that 1≤i<j≤l1≤i<j≤l and ai<ajai<aj. For example, the sequence [0,2,0,2,0][0,2,0,2,0] has an ascent because of the pair (1,4)(1,4), but the sequence [4,3,3,3,1][4,3,3,3,1] doesn’t have an ascent.

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

Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has nn sequences s1,s2,…,sns1,s2,…,sn which may have different lengths.

Gyeonggeun will consider all n2n2 pairs of sequences sxsx and sysy (1≤x,y≤n1≤x,y≤n), and will check if its concatenation sx+sysx+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,yx,y) of sequences s1,s2,…,sns1,s2,…,sn whose concatenation sx+sysx+sy contains an ascent.

Input
The first line contains the number nn (1≤n≤1000001≤n≤100000) denoting the number of sequences.

The next nn lines contain the number lili (1≤li1≤li) denoting the length of sisi, followed by lili integers si,1,si,2,…,si,lisi,1,si,2,…,si,li (0≤si,j≤1060≤si,j≤106) denoting the sequence sisi.

It is guaranteed that the sum of all lili does not exceed 100000100000.

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

Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
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
Output
72
Note
For the first example, the following 99 arrays have an ascent: [1,2],[1,2],[1,3],[1,3],[1,4],[1,4],[2,3],[2,4],[3,4][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：给定n个数组，按照数组加法那样把任意两个数组拼接到一起。如果有1≤i<j≤l且ai <aj这样的，就是一个题目要求的连接串，也可以自己和自己连接。问一共可以有多少个连接串。
思路：如果数组本身就是这样的连接串，那么它和任意一个数组都可以组成题目要求的连接串。如果数组ai不是题目中的连接串，并且这个数组的最大值比其余数组aj的最小值还要小的话，这样就不能组成连接串了（ai+aj不行，但是aj+ai可以）。那么我们可以把所有的不是题目要求的连接串中的最大值最小值求出来，然后对最小值数组排序，之后把最大值数组去最小值数组中二分，找出不符合的，最后用n*n减去就行。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxx=1e5+100;
vector<int> p[maxx];
int l[maxx],r[maxx],vis[maxx];
int n;int main()
{scanf("%d",&n);int x,y;for(int i=1;i<=n;i++){scanf("%d",&x);for(int j=0;j<x;j++){scanf("%d",&y);p[i].push_back(y);}}memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++){for(int j=0;j<p[i].size();j++){for(int k=j+1;k<p[i].size();k++){if(p[i][j]<p[i][k]){vis[i]=1;break;}}if(vis[i]) break;}}int cnt=0;for(int i=1;i<=n;i++){if(!vis[i]){sort(p[i].begin(),p[i].end());l[++cnt]=p[i][0],r[cnt]=p[i][p[i].size()-1];}}ll num=0;sort(l+1,l+1+cnt);for(int i=1;i<=cnt;i++){int pos=lower_bound(l+1,l+1+cnt,r[i])-l-1;num+=cnt-pos;}cout<<((ll)n*(ll)n)-num<<endl;//注意用long long
}

努力加油a啊，(^o)/~

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