hdu 2150(判断折线是否相交)

 1 /*

 2 *  题目要求：判断给出的一些折线段是否相交 

 3 *  解法：枚举每条折线段上的每段线段，判断其是否与其它折线段的线段相交 

 4 */

 5 

 6 #include <cstdio>

 7 #include <cstdlib>

 8 #include <iostream>

 9 

10 using  namespace std;

11 

12 const int N = 35;

13 const int M = 105;

14 

15 int num[N];

16 struct point {

17     double x;

18     double y;

19 }p[N][M];

20 

21 double crossProd(point A, point B, point C) {

22     return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);

23 }

24 

25 bool segIntersect(point A, point B, point C, point D) {

26     if (max(A.x, B.x) >= min(C.x, D.x) &&

27         max(C.x, D.x) >= min(A.x, B.x) &&

28         max(A.y, B.y) >= min(C.y, D.y) &&

29         max(C.y, D.y) >= min(A.y, B.y) &&

30         crossProd(A, B, C)*crossProd(A, D, B) > 0 &&

31         crossProd(C, D, A)*crossProd(C, B, D) > 0) return true;

32     return false;

33 }

34 

35 bool solve(int n) {

36     for (int i=0; i<n-1; ++i) {//枚举每条折线段 

37         for (int j=1; j<num[i]; ++j) {

38             for (int ii=i+1; ii<n; ++ii) {

39                 for (int k=1; k<num[ii]; ++k) {

40                     if (segIntersect(p[i][j-1], p[i][j], p[ii][k-1], p[ii][k])) return false;//判断线段相交 

41                 }

42             }

43         }

44     }

45     return true;

46 }

47 

48 int main() {

49     int n;

50     while (scanf("%d", &n) != EOF) {

51         for (int i=0; i<n; ++i) {

52             scanf ("%d", &num[i]);

53             for (int j=0; j<num[i]; ++j) scanf ("%lf%lf", &p[i][j].x, &p[i][j].y);

54         }

55         if (n == 1) printf ("No\n");

56         else {

57             bool yes = solve(n);

58             if (yes) printf ("No\n");

59             else printf ("Yes\n");

60         }    

61     }

62     return 0;

63 }