hdu 2150(判断折线是否相交)

/*
*  题目要求：判断给出的一些折线段是否相交 
*  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 
*/

#include <cstdio>
#include <cstdlib>
#include <iostream>

using  namespace std;

const int N = 35;
const int M = 105;

int num[N];
struct point {
    double x;
    double y;
}p[N][M];

double crossProd(point A, point B, point C) {
    return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
}

bool segIntersect(point A, point B, point C, point D) {
    if (max(A.x, B.x) >= min(C.x, D.x) &&
        max(C.x, D.x) >= min(A.x, B.x) &&
        max(A.y, B.y) >= min(C.y, D.y) &&
        max(C.y, D.y) >= min(A.y, B.y) &&
        crossProd(A, B, C)*crossProd(A, D, B) > 0 &&
        crossProd(C, D, A)*crossProd(C, B, D) > 0) return true;
    return false;
}

bool solve(int n) {
    for (int i=0; i<n-1; ++i) {//枚举每条折线段 
        for (int j=1; j<num[i]; ++j) {
            for (int ii=i+1; ii<n; ++ii) {
                for (int k=1; k<num[ii]; ++k) {
                    if (segIntersect(p[i][j-1], p[i][j], p[ii][k-1], p[ii][k])) return false;//判断线段相交 
                }
            }
        }
    }
    return true;
}

int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        for (int i=0; i<n; ++i) {
            scanf ("%d", &num[i]);
            for (int j=0; j<num[i]; ++j) scanf ("%lf%lf", &p[i][j].x, &p[i][j].y);
        }
        if (n == 1) printf ("No\n");
        else {
            bool yes = solve(n);
            if (yes) printf ("No\n");
            else printf ("Yes\n");
        }    
    }
    return 0;
}