hdu 2907 凹陷面

来源：互联网发布：c语言控制机械手编辑：程序博客网时间：2024/04/28 12:22

给定一个多边形，让你求出其价值。价值的定义是：-p*凹面的个数+q*凸面的个数。。。最小值为0

凸面的个数就是凸包中的点的个数，但是当出现凹面时，就会减少一个凸面，这是因为这时候的凸面是虚拟出来的！！！！！

题意参考这里点击打开链接

int  cmp(double x){     if(fabs(x) < 1e-8) return 0 ;     if(x > 0) return 1 ;     return -1  ;}struct point{       double  x ,  y  ;       int  id ;       point(){}       point(double _x , double _y):x(_x) , y(_y){}       friend bool operator == (const point &a , const point &b){              return cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0 ;       }       friend double operator ^ (const point &a , const point &b){              return a.x * b.y - a.y * b.x ;       }       friend double dist(const  point &a , const point &b){              return  sqrt( (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y) ) ;       }       point operator - (point o){             return  point(x - o.x , y - o.y) ;       }};bool  cmpless(const  point &a , const point &b){      return   cmp(a.x - b.x) < 0            || cmp(a.x - b.x == 0) && cmp(a.y - b.y) < 0 ;}vector<point> convex_hull(vector<point> a){     vector<point> src(2 * a.size() + 5)  ;     sort(a.begin() , a.end() , cmpless) ;     a.erase(unique(a.begin() , a.end()) , a.end())  ;     int m = 0 ;     for(int i = 0 ; i < a.size() ; i++){          while(m > 1 && cmp( (src[m-1] - src[m-2]) ^  (a[i] - src[m-2]) ) <= 0)              m-- ;          src[m++] = a[i] ;     }     int k = m ;     for(int i = a.size() - 2 ; i >= 0 ; i--){          while(m > k && cmp( (src[m-1] - src[m-2]) ^ (a[i] - src[m-2])) <= 0)              m-- ;          src[m++] = a[i] ;     }     src.resize(m) ;     if(a.size() > 1) src.resize(m-1) ;     return src ;}int   vis[38] ;int   main(){      int n  ,  i  , t  , pcnt , qcnt , p , q ;      cin>>t  ;      while(t--){           cin>>p>>q>>n ;           vector<point> lis(n) ;           for(i = 0 ; i < n ; i++){                scanf("%lf%lf" , &lis[i].x , &lis[i].y) ;                lis[i].id = i ;           }           vector<point> hull = convex_hull(lis) ;           qcnt = hull.size() ;           memset(vis , 0 , sizeof(vis)) ;           for(i = 0 ; i < hull.size() ; i++) vis[hull[i].id] = 1 ;           pcnt = 0 ;           for(i = 0 ; i < n ; i++){                if(vis[i] && !vis[(i+1)%n]) pcnt++ ;           }           qcnt -= pcnt ;           cout<< max(0 , q * qcnt - p * pcnt )<< endl ;      }      return  0  ;}

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