2016-2017 National Taiwan University World Final Team Selection Contest C

题目：
Statements

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：
　　直接通过枚举y来判断直线经过哪些方格，vis标记下即可。
　　这题好像精度挺重要的。。。。

#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-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int vis[3001][3001];
struct Point
{
    double x,y;
};
void solve(void)
{
    Point st,se;
    scanf("%lf%lf%lf%lf",&st.x,&st.y,&se.x,&se.y);
    if(fabs(st.x-se.x)<eps||fabs(st.y-se.y)<eps) return;
    if(st.y>se.y) swap(st,se);
    double k=(se.x-st.x)/(se.y-st.y),b=st.x-k*st.y;
    double tmp;
    if(k>0)
    {
        for(int i=se.y-1,mx,mi;i>=st.y;i--)
        {
            tmp=(i+1)*k+b;
            mi=floor(i*k+b);
            if(fabs(tmp-floor(tmp))<=eps)mx=max(0,(int)tmp-1);
            else mx=(int)tmp;
            if(mi<0) while(1);
            while(mi<=mx)
                vis[mi++][i]=1;
        }
    }
    else
    {
        for(int i=st.y+1,mx,mi;i<=se.y;i++)
        {
            tmp=(i-1)*k+b;
            mi=floor(i*k+b);
            if(fabs(tmp-floor(tmp))<=eps)mx=max(0,(int)tmp-1);
            else mx=(int)tmp;
            if(mi<0) while(1);
            while(mi<=mx)
                vis[mi++][i-1]=1;
        }
    }
}
int main(void)
{
    int n,cnt=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        solve();
    for(int i=0;i<2000;i++)
    for(int j=0;j<2000;j++)
        cnt+=vis[i][j];
    printf("%d
",cnt);
    return 0;
}