题目链接：http://codeforces.com/problemset/problem/216/D

题意：

对于一个梯形区域，假设梯形左边的点数！=梯形右边的点数，那么这个梯形为红色。否则为绿色，

问：

给定的蜘蛛网中有多少个红色。

2个树状数组维护2个线段。然后暴力模拟一下，由于点数非常多但须要用到的线段树仅仅有3条，所以类似滚动数组的思想优化内存。

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<algorithm>

#include<vector>

#include<set>

using namespace std;

#define N 10010

#define L(x) (x<<1)

#define R(x) (x<<1|1)

#define ll int

int maxn;

struct hehe{

	int c[100505];

	void init(){memset(c, 0, sizeof c);}

	inline int Lowbit(int x){return x&(-x);}

	void change(int i, int x)//i点增量为x

	{

		while(i <= maxn)

		{

			c[i] += x;

			i += Lowbit(i);

		}

	}

	int sum(int x){//区间求和 [1,x]

		int ans = 0;

		for(int i = x; i >= 1; i -= Lowbit(i))

			ans += c[i];

		return ans;

	}

}tree[2];

int ans;

vector<int>l,r,o;

void work(){

	if(o.size()<=1)return;

	tree[0].init(); tree[1].init();

	for(int i = 0; i < l.size(); i++)

		tree[0].change(l[i],1);

	for(int i = 0; i < r.size(); i++)

		tree[1].change(r[i],1);

	int L = o[0];

	for(int i = 1; i < o.size(); i++){

		int R = o[i];

		if(L+1<=R-1){

			int Z = tree[0].sum(R-1)-tree[0].sum(L);

			int Y = tree[1].sum(R-1)-tree[1].sum(L);

			ans += (Z!=Y);

		}

		L = R;

	}

}

vector<int>a,tmp1, red, tmp2, tmpend;

void Red(){

	red.clear();

	int HHH,EEE; scanf("%d",&HHH);

	while(HHH--){scanf("%d",&EEE);red.push_back(EEE);}

	sort(red.begin(),red.end());

}

int n;

int main(){

	int i,j,num;

	maxn = 100010;

	while(~scanf("%d",&n)){

		ans = 0;

		l.clear(); r.clear(); tmp1.clear(); a.clear();

		tmp2.clear(); tmpend.clear();

		Red();

		l = tmp1 = red;

		Red();

		tmp2 = o = red;

		for(i = 3; i <= n; i++){

			Red();

			r = red;

			if(i==n)tmpend = red;

			work();

			l = o;

			o = r;

		}

		r = tmp1;

		work();

		l = tmpend;

		o = tmp1;

		r = tmp2;

		work();

		cout<<ans<<endl;

	}

	return 0;

}

/*

3

2 1 3

3 1 3 2

3 1 3 2

ans:

0

2

*/

/*

ll n,m,k;

ll a[N];

ll gcd(ll x,ll y){

if(x>y)swap(x,y);

while(x){

y%=x;

swap(x,y);

}

return y;

}

int main(){

ll i, j, u, v, que;

while(cin>>n>>m>>k){

for(i = 1; i <= n; i++)cin>>a[i];

ll _gcd = a[1];

for(i = 2; i <= n; i++)_gcd = gcd(_gcd,a[i]);

for(i = 1; i <= n; i++)a[i]/=_gcd;

}

return 0;

}

/*

*/