Computer College Undergraduate Program Design Contest (2015 ') Polygon

Polygon Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 195 accepted Submission (s): 41


Problem Description give you a simple polygon and a line, and your task are to calculate the length of the line which is Co Vered by the polygon.
Input There are several test cases.
Each test case starts with one non-negative integer n (3 <= n <= 1000) giving the number of the vertices of the poly Gon. Then following n lines, each line contains two real numbers standing for the x and y coordinates of a vertex. The vertices are given either in clockwise or counterclockwise order. Then four real numbers (SX, SY, TX, Ty), standing for the coordinates of the two points of the.
All real numbers are between-100000 and 100000.
Output for each case print the "total length" of the line this covered by the polygon, with 3 digits on the decimal point .

Sample Input

The 4 0 0 0 1 1 1 1 0 0 0 1 1 4 0 0 0 1--------------+ 1 1 1 0 0, 0
Sample Output

1.414 1.000 6.000

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#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <
Algorithm> using namespace std;
    struct point {double x;
Double y;
} p[1001], px[10001];
int n;
Double eps=1e-8;
    int CMP (point A, point B) {if (ABS (a.x-b.x) <eps) return a.y<b.y;
Return a.x<b.x; Double Dist (point A,point b) {return sqrt (a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y)), double direction (point Pi,p Oint pj,point PK) {return (pk.x-pi.x) * (PJ.Y-PI.Y)-(pj.x-pi.x) * (PK.Y-PI.Y);} bool On_segment (Point Pi,point pj,point PK) {if (Direction (pi, PJ, pk) ==0) {if (Pk.x>=min (pi.x,pj.x) &&pk.x<=max (pi.x,pj.x) &&
    Pk.y>=min (PI.Y,PJ.Y) &&pk.y<=max (PI.Y,PJ.Y)) return true;
return false;
    BOOL Segments_intersect (Point P1,point p2,point p3,point p4) {double d1=direction (P3,P4,P1);
    Double d2=direction (P3,P4,P2);
    Double d3=direction (P1,P2,P3); Double D4=directiOn (P1,P2,P4);
    if (d1*d2<0&&d3*d4<0) return true;
    else if (d1==0&&on_segment (P3,P4,P1)) return true;
    else if (d2==0&&on_segment (P3,P4,P2)) return true;
    else if (d3==0&&on_segment (P1,P2,P3)) return true;
    else if (d4==0&&on_segment (P1,P2,P4)) return true;
return false;
    Point intersection (Point A1, Point A2, point B1, point B2) {point ret = A1;  Double T = ((a1.x-b1.x) * (B1.Y-B2.Y)-(A1.Y-B1.Y) * (b1.x-b2.x))/((a1.x-a2.x) * (B1.Y-B2.Y)
    -(A1.Y-A2.Y) * (b1.x-b2.x));
    Ret.x + + (a2.x-a1.x) * t;
    Ret.y + + (a2.y-a1.y) * t;
return ret;
    int Inpolygon (point a) {int i;
    Point B,c,d;
    B.Y=A.Y;
    B.X=1E15;
    int count=0;
        For (i=0 i<n; i++) {c = p[i];
        D = p[i + 1];
        if (On_segment (C,d,a)) return 1;
        if (ABS (C.Y-D.Y) <eps) continue; if (On_segment (a,b,c) {if (c.y>d.y) count++;
        else if (on_segment (a,b,d)) {if (d.y>c.y) count++;
    else if (Segments_intersect (a,b,c,d)) count++;
return count%2; BOOL Intersect (point s,point e,point a,point b) {return direction (e,a,s) *direction (e,b,s) <=0;} double Calculate
(Point S,point E)
    {int i,k=0;
    Double sum;
    Point A,b,temp;
        for (i=0; i<n; i++) {a=p[i];
        B=P[I+1];
            if (ABS (direction (e,a,s)) <eps&&abs (direction (e,b,s)) <eps) {px[k++]=a;
        Px[k++]=b;
        else if (Intersect (s,e,a,b)) {px[k++]=intersection (s,e,a,b);
    } if (k==0) return 0.0;
    Sort (px,px+k,cmp);
    PX[K]=PX[0];
    sum=0;
        for (i=0; i<k-1; i++) {a=px[i];
        B=PX[I+1];
        temp.x= (a.x+b.x)/2.0;
   temp.y= (A.Y+B.Y)/2.0;     if (Inpolygon (temp)) sum+=dist (A,B);
return sum;
    int main () {int i;
    Double sum;
    Point S,e; while (~SCANF ("%d", &n) &&n) {for (i=0; i<n; i++) scanf ("%lf%lf", &p[i].x,&p[i)
        . y);
        P[N]=P[0];
        scanf ("%lf%lf%lf%lf", &s.x,&s.y,&e.x,&e.y);
        Sum=calculate (s,e);
    printf ("%.3lf\n", sum);
return 0;
 }

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for what this time the topic is in English .... Qaq ...