uva10256 The Great Divide（凸包+判断）

uva10256

题目

就是给你一些点，一些点支持点A，另一些支持点B，现在问能不能画一条直线使得两个阵营的人正好在两边。

思路

还是凸包，但是注意一个点和两个点的情况，同时注意判断一个凸包在另一个凸包里面的情况，其他时候就暴力判断直线是否相交即可。

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define maxn 511#define pi acos (-1)#define rotate Rotateusing namespace std;typedef long long ll;const double eps = 1e-8;int dcmp (double x){    if (fabs (x) < eps)        return 0;    else return x < 0 ? -1 : 1;}struct point{    double x, y;    point (double _x = 0, double _y = 0) : x(_x), y(_y) {}    point operator - (point a) const    {        return point (x-a.x, y-a.y);    }    point operator + (point a) const    {        return point (x+a.x, y+a.y);    }    bool operator < (const point &a) const    {        return x < a.x || (x == a.x && y < a.y);    }    bool operator == (const point &a) const    {        return dcmp (x-a.x) == 0 && dcmp (y-a.y) == 0;    }};point operator * (point a, double p){    return point (a.x*p, a.y*p);}double cross (point a, point b){    return a.x*b.y-a.y*b.x;}double dot (point a, point b){    return a.x*b.x + a.y*b.y;}int ConvexHull (point *p, point *ch, int n){    sort (p, p+n);    int m = 0;    for (int i = 0; i < n; i++)    {        while (m > 1 && cross (ch[m-1]-ch[m-2], p[i]-ch[m-1]) <= 0)            m--;        ch[m++] = p[i];    }    int k = m;    for (int i = n-2; i >= 0; i--)    {        while (m > k && cross (ch[m-1]-ch[m-2], p[i]-ch[m-1]) <= 0)            m--;        ch[m++] = p[i];    }    if (n > 1)        m--;    return m;}bool OnSegment (point p, point a1, point a2)   //点在线段上{    return dcmp (cross (a1-p, a2-p)) == 0 && dcmp (dot (a1-p, a2-p)) <= 0;}bool SegmentIntersection (point a1, point a2, point b1, point b2)   //线段相交{    if (OnSegment (a1, b1, b2) || OnSegment (a2, b1, b2) || OnSegment (b1, a1, a2) || OnSegment (b2, a1, a2))        return 1;    double c1 = cross (a2-a1, b1-a1), c2 = cross (a2-a1, b2-a1), c3 = cross (b2-b1, a1-b1),           c4 = cross (b2-b1, a2-b1);    return dcmp (c1)*dcmp (c2) < 0 && dcmp (c3)*dcmp (c4) < 0;}int PointInPolygon (point p, point *poly, int n){    int wn = 0;    for (int i = 0; i < n; i++)    {        if (OnSegment (p, poly[i], poly[(i+1)%n]))            return -1;        int k = dcmp (cross (poly[(i+1)%n]-poly[i], p-poly[i]));        int d1 = dcmp (poly[i].y-p.y);        int d2 = dcmp (poly[(i+1)%n].y-p.y);        if (k > 0 && d1 <= 0 && d2 > 0) wn++;        if (k < 0 && d2 <= 0 && d1 > 0) wn--;    }    if (wn != 0)        return 1;    return 0;}int n,m,cnt1,cnt2;point p1[maxn],p2[maxn],ch1[maxn],ch2[maxn];bool check(){    if(cnt1==1)    {        if(cnt2==1&&ch1[0]==ch2[0])            return 0;        else if(cnt2==2&&OnSegment(ch1[0],ch2[0],ch2[1]))            return 0;        else if(PointInPolygon(ch1[0],ch2,cnt2))            return 0;    }    else if(cnt1==2)    {        if(cnt2==1&&OnSegment(ch2[0],ch1[0],ch1[1]))            return 0;        else if(cnt2==2&&SegmentIntersection(ch1[0],ch1[1],ch2[0],ch2[1]))            return 0;    }    if (PointInPolygon (ch2[0], ch1, cnt1) || PointInPolygon (ch1[0], ch2, cnt2))        return 0;    for (int i = 0; i < cnt1; i++)    {        for (int j = 0; j < cnt2; j++)        {            if (SegmentIntersection (ch1[i], ch1[(i+1)%cnt1], ch2[j], ch2[(j+1)%cnt2]))                return 0;        }    }    return 1;}int main(){    while(scanf("%d %d",&n,&m)!=EOF&&(n+m))    {        for(int i=0; i<n; i++)            scanf("%lf %lf",&p1[i].x,&p1[i].y);        for(int i=0; i<m; i++)            scanf("%lf %lf",&p2[i].x,&p2[i].y);        cnt1=ConvexHull(p1,ch1,n);        cnt2=ConvexHull(p2,ch2,m);        if (check ())            printf ("Yes\n");        else printf ("No\n");    }    return 0;}