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：
codeforces题解哪里看。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 2ans:
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;
}
/**/
