HDU 1392
凸包模板题
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cmath> 5 using namespace std; 6 int n,top; 7 struct point{ 8 double x,y; 9 }tree[105],sta[105]; 10 double dis(point a,point b){ 11 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 12 } 13 double multiply(point a,point b,point c){ 14 return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x); 15 } 16 bool cmp(point a,point b){ 17 return a.x==b.x?a.y<b.y:a.x<b.y; 18 } 19 int cmp ( const void *p1 , const void *p2 ) 20 { 21 point *p3,*p4; 22 double m; 23 p3 = (point *)p1; 24 p4 = (point *)p2; 25 m = multiply(tree[0] , *p3 , *p4) ; 26 if(m < 0) return 1; 27 else if(m == 0 && (dis(tree[0] , *p3) < dis(tree[0],*p4))) 28 return 1; 29 else return -1; 30 } 31 void Graham(){ 32 int i ; 33 sta[0].x = tree[0].x; 34 sta[0].y = tree[0].y; 35 sta[1].x = tree[1].x; 36 sta[1].y = tree[1].y; 37 sta[2].x = tree[2].x; 38 sta[2].y = tree[2].y; 39 top = 2; 40 for ( i = 3 ; i <= n ; ++ i ) 41 { 42 while (multiply(sta[top - 1] , sta[top] , tree[i]) <= 0 ) 43 top -- ; //出栈 44 sta[top + 1].x = tree[i].x ; 45 sta[top + 1].y = tree[i].y ; 46 top ++ ; 47 } 48 } 49 int main(){ 50 int pos; 51 double temp,px,py; 52 while(scanf("%d",&n)&&n){ 53 for(int i=0;i<n;i++){ 54 scanf("%lf%lf",&tree[i].x,&tree[i].y); 55 } 56 if(n==1){printf("0.00\n");continue;} 57 else if(n==2){printf("%.2lf\n",dis(tree[0],tree[1]));continue;} 58 px=tree[0].x,py=tree[0].y,pos=0; 59 for(int i=1;i<n;i++){ 60 if(tree[i].x<px){ 61 px=tree[i].x; 62 py=tree[i].y; 63 pos=i; 64 } 65 else if(tree[i].x==px&&tree[i].y<py){ 66 px=tree[i].x; 67 py=tree[i].y; 68 pos=i; 69 } 70 } 71 temp = tree[0].x ; // 找出左下角的点 72 tree[0].x = tree[pos].x ; 73 tree[pos].x = temp ; 74 temp = tree[0].y ; 75 tree[0].y = tree[pos].y ; 76 tree[pos].y = temp ; 77 qsort(&tree[1],n - 1,sizeof(double) * 2,cmp); 78 //for(int i=0;i<n;i++)printf("%lf %lf\n",tree[i].x,tree[i].y); 79 tree[n].x=tree[0].x,tree[n].y=tree[0].y; 80 Graham(); 81 double len = 0.0; 82 for(int i = 0 ; i < top ; i ++) 83 len = len + dis(sta[i] , sta[i+1]) ; 84 printf("%.2lf\n",len); 85 } 86 return 0; 87 }
