2016-2017 National Taiwan University World Final Team Selection Contest A - Hacker Cups and Balls

题目：

Dreamoon likes algorithm competitions very much. But when he feels crazy because he cannot figure out any solution for any problem in a competition, he often draws many meaningless straight line segments on his calculation paper.

Dreamoon's calculation paper is special: it can be imagined as the plane with Cartesian coordinate system with range [0, 2000] × [0, 2000] for the coordinates. The grid lines are all lines of the form x = c or y = c for every integer c between 0 and 2000, inclusive. So, the grid contains 2000 × 2000 squares.

Now, Dreamoon wonders how many grid squares are crossed by at least one of the lines he drew. Please help Dreamoon find the answer. Note that, to cross a square, a segment must contain an interior point of the square.

Input

The first line of input contains an integer N denoting the number of lines Dreamoon draw. The i-th line of following N lines contains four integers x_i1, y_i1, x_i2, y_i2, denoting that the i-th segment Dreamoon drew is a straight line segment between points (x_i1, y_i1) and (x_i2, y_i2).

• 1 ≤ N ≤ 2 × 10³
• 0 ≤ x_i1, y_i1, x_i2, y_i2 ≤ 2 × 10³
• the lengths of all line segments in input are non-zero

Output

Output one integer on a single line: how many grid squares are crossed by at least one of the line segments which Dreamoon drew.

Example

Input

3
0 0 5 5
0 5 5 0
0 5 5 0

Output

9

Input

1
0 0 4 3

Output

6

Input

2
0 0 4 3
1 0 3 3

Output

6

思路：

　　二分答案，check时，把小于等于二分值的数标记为0，大于的标记为1。

　　然后排序就是区间内的0和1进行排序，用线段树维护区间0的个数的话，排序就是把一段区间置为0或1。

　　最后根据center为是不是为0.

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=1e6+;
 const int mod=1e9+;

 struct node
 {
     int l,r;
 }q[K];
 int n,m,a[K],sum[K],ff[K];
 void push_down(int o,int l,int r)
 {
     int mid=l+r>>;
     if(ff[o]&&sum[o])
         sum[o<<]=mid-l+,sum[o<<|]=r-mid;
     else if(ff[o])
         sum[o<<]=,sum[o<<|]=;
     if(ff[o])
         ff[o<<]=ff[o<<|]=;
     ff[o]=;
 }
 void build(int o,int l,int r,int v)
 {
     ff[o]=;
     if(l==r)
     {
         if(a[l]<=v) sum[o]=;
         else sum[o]=;
         return ;
     }
     build(o<<,l,l+r>>,v),build(o<<|,(l+r)/+,r,v);
     sum[o]=sum[o<<]+sum[o<<|];
 }
 void update(int o,int l,int r,int nl,int nr,int x)
 {
     //if(nl>nr) while(1);
     if(l==nl&&r==nr)
     {
         if(x==)    sum[o]=r-l+;
         else sum[o]=;
         ff[o]=;
         return ;
     }
     push_down(o,l,r);
     int mid=l+r>>;
     if(nr<=mid) update(o<<,l,mid,nl,nr,x);
     else if(nl>mid) update(o<<|,mid+,r,nl,nr,x);
     else update(o<<,l,mid,nl,mid,x),update(o<<|,mid+,r,mid+,nr,x);
     sum[o]=sum[o<<]+sum[o<<|];
 }
 int query(int o,int l,int r,int nl,int nr)
 {
     //if(nl>nr) while(1);
     if(l==nl&&r==nr)    return sum[o];
     push_down(o,l,r);
     int mid=l+r>>;
     if(nr<=mid) return query(o<<,l,mid,nl,nr);
     else if(nl>mid) return query(o<<|,mid+,r,nl,nr);
     return query(o<<,l,mid,nl,mid)+query(o<<|,mid+,r,mid+,nr);
 }
 bool check(int x)
 {
     build(,,n,x);
     for(int i=;i<=m;i++)
     if(q[i].l<q[i].r)
     {
         int cnt=query(,,n,q[i].l,q[i].r);
         if(cnt)
             update(,,n,q[i].l,q[i].l+cnt-,);
         if(q[i].l+cnt<=q[i].r)
             update(,,n,q[i].l+cnt,q[i].r,);
     }
     else if(q[i].l>q[i].r)
     {
         int cnt=q[i].l-q[i].r+-query(,,n,q[i].r,q[i].l);
         if(cnt)
         update(,,n,q[i].r,q[i].r+cnt-,);
         if(q[i].r+cnt<=q[i].l)
             update(,,n,q[i].r+cnt,q[i].l,);
     }
     return query(,,n,(n+)/,(n+)/)==;
 }

 int main(void)
 {
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)
         scanf("%d",a+i);
     for(int i=;i<=m;i++)
         scanf("%d%d",&q[i].l,&q[i].r);
     int l=,r=n,ans=;
     while(l<=r)
     {
         int mid=l+r>>;
         if(check(mid)) r=mid-,ans=mid;
         else l=mid+;
     }
     printf("%d\n",ans);
     return ;
 }