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  • 求两条线段的交点

      两条线段的两个端点坐标(x1,y1) (x2,y2) (x3,y3) (x4,y4)

      b1=(y2-y1)*x1+(x1-x2)*y1

      b2=(y4-y3)*x3+(x3-x4)*y3

      D=(x2-x1)(y4-y3)-(x4-x3)(y2-y1)

      D1=b2*(x2-x1)-b1*(x4-x3)

      D2=b2*(y2-y1)-b1*(y4-y3)

      交点(x0,y0)

      x0=D1/D   y0=D2/D

    推导:http://www.cnblogs.com/dwdxdy/p/3230485.html

    E. Covered Points
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You are given nn segments on a Cartesian plane. Each segment's endpoints have integer coordinates. Segments can intersect with each other. No two segments lie on the same line.

    Count the number of distinct points with integer coordinates, which are covered by at least one segment.

    Input

    The first line contains a single integer nn (1n10001≤n≤1000) — the number of segments.

    Each of the next nn lines contains four integers Axi,Ayi,Bxi,ByiAxi,Ayi,Bxi,Byi (106Axi,Ayi,Bxi,Byi106−106≤Axi,Ayi,Bxi,Byi≤106) — the coordinates of the endpoints AA, BB (ABA≠B) of the ii-th segment.

    It is guaranteed that no two segments lie on the same line.

    Output

    Print a single integer — the number of distinct points with integer coordinates, which are covered by at least one segment.

    Examples
    input
    9
    0 0 4 4
    -1 5 4 0
    4 0 4 4
    5 2 11 2
    6 1 6 7
    5 6 11 6
    10 1 10 7
    7 0 9 8
    10 -1 11 -1
    output
    42
    input
    4
    -1 2 1 2
    -1 0 1 0
    -1 0 0 3
    0 3 1 0
    output
    7

    The image for the first example:

    Several key points are marked blue, the answer contains some non-marked points as well.

    The image for the second example:

    求线段进过的整数点。线段经过的点,为x轴长度,与y轴长度的gcd。
    #include<bits/stdc++.h>
    #define ll long long
    using namespace std;
    
    struct Point{
        ll x,y;
        Point(ll x=0,ll y=0):x(x),y(y){};
    };
    
    ll gcd(ll a,ll b)
    {
        return a==0?b:gcd(b%a,a);
    }
    
    bool cheak(ll op1,ll a,ll b){
        if(a>b)
            swap(a,b);
        return op1>=a&&op1<=b;
    }
    
    Point point_of_intersection(Point f1,Point f2,Point f3,Point f4,bool &mark)
    {
        ll a1,a2,b1,b2,c1,c2,c3,c4,D,D1,D2;
        a1=f2.y-f1.y;
        a2=f1.x-f2.x;
        b1=a1*f1.x+a2*f1.y;
        ///b1=(y2-y1)*x1+(x1-x2)*y1
        c1=f4.y-f3.y;
        c2=f3.x-f4.x;
        b2=c1*f3.x+c2*f3.y;
        ///b2=(y4-y3)*x3+(x3-x4)*y3
        c3=f2.x-f1.x;
        c4=f4.x-f3.x;
    
        D=c3*c1-c4*a1;
        Point res;
        if(D==0){
            mark=false;
            return res;
        }
        D1=b2*c3-b1*c4;
        if(D1%D){
            mark=false;
            return res;
        }
        res.x=int(D1/D);
    
        D2=b2*a1-b1*c1;
        if(D2%D){
            mark=false;
            return res;
        }
        res.y=int(D2/D);
    
        if(!cheak(res.x,f1.x,f2.x)||!cheak(res.x,f3.x,f4.x)){
            mark=false; return res;
        }
        if(!cheak(res.y,f1.y,f2.y)||!cheak(res.y,f3.y,f4.y)){
            mark=false; return res;
        }
        return res;
    }
    
    Point edge[1006][2];
    int main()
    {
        int n;
        while( ~scanf("%d",&n)){
            for(int i=1;i<=n;i++){
                scanf("%lld%lld%lld%lld",&edge[i][0].x,&edge[i][0].y,&edge[i][1].x,&edge[i][1].y);
            }
    
            set<pair<ll,ll> > re;
            long long ans=0,tmp;
            bool mark;
            for(int i=1;i<=n;i++){
                tmp=gcd(abs(edge[i][0].x-edge[i][1].x),abs(edge[i][0].y-edge[i][1].y))+1;
                re.clear();
                for(int j=1;j<i;j++){
                    mark=true;
                    Point res=point_of_intersection(edge[i][0],edge[i][1],edge[j][0],edge[j][1],mark);
                    if(mark)
                        re.insert(make_pair(res.x,res.y));
                }
                ans+=tmp-re.size();
            }
            printf("%lld
    ",ans);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/ZQUACM-875180305/p/10149769.html
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