凸包问题
poj 1113 http://poj.org/problem?id=1113
题意就是给你一些点，然后让你求凸包并求出这个凸包的周长，由于题目中有要求，所以求出周长后还要再加上一个以输入 L 为半径的园的周长，才是所求的答案
View Code
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <stdlib.h> 6 #include <math.h> 7 #define N 1010 8 #define pi 3.1415926 9 10 using namespace std; 11 12 struct node 13 { 14 double x; 15 double y; 16 }point[N],stack[N]; 17 int n,top,l; 18 double dis(node a,node b) 19 { 20 return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); 21 } 22 double mul(node &a,node &b,node &c) 23 { 24 return (c.x - a.x) * (b.y - a.y) - (b.x - a.x) * (c.y - a.y); 25 } 26 /* 极角排序 */ 27 int cmp(const void *a,const void *b) 28 { 29 node *c = (node *) a; 30 node *d = (node *) b; 31 int m = mul(point[0],*c,*d); 32 if(m == 0) return dis(*c,point[0]) - dis(*d,point[0]); 33 else return -m; 34 } 35 /* */ 36 void gramham() 37 { 38 top = 1; 39 qsort(point + 1,n - 1,sizeof(node),cmp); 40 stack[0] = point[0]; 41 stack[1] = point[1]; 42 int i; 43 for(i = 2; i < n; i++) 44 { 45 while(top >= 1 && mul(stack[top - 1],stack[top],point[i]) <= 0) // 根据叉积判断是左转还是右转 46 top --; 47 stack[++top] = point[i]; 48 } 49 } 50 int main() 51 { 52 int i; 53 int kemp; 54 double maxx; 55 //freopen("data.txt","r",stdin); 56 while(cin>>n>>l) 57 { 58 memset(point,0,sizeof(point)); 59 memset(stack,0,sizeof(stack)); 60 kemp = 0; 61 for(i = 0; i < n; i++) 62 { 63 scanf("%lf%lf",&point[i].x,&point[i].y); 64 if(point[i].y < point[kemp].y || (point[i].y == point[kemp].y && point[i].x < point[kemp].x)) // 查找最靠下，靠左的点 65 kemp = i; 66 } 67 swap(point[0],point[kemp]); 68 gramham(); 69 stack[top + 1] = stack[0]; 70 maxx = 0; 71 for(i = 0; i < top + 1; i++) 72 { 73 maxx += (dis(stack[i],stack[i + 1]) * 1.0); 74 } 75 maxx += pi * 2 * l; 76 printf("%.0lf\n",maxx); 77 } 78 return 0; 79 }
poj 2187
题意：利用凸包求最远距离，与上题方法一样不多解释了
View Code
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <stdlib.h> 6 #define N 50010 7 8 using namespace std; 9 10 struct node 11 { 12 int x; 13 int y; 14 }point[N],stack[N]; 15 int n,top; 16 int dis(node &a,node &b) 17 { 18 return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y); 19 } 20 int mul(node &a,node &b,node &c) 21 { 22 return (c.x - a.x) * (b.y - a.y) - (b.x - a.x) * (c.y - a.y); 23 } 24 int cmp(const void *a,const void *b) 25 { 26 node *c = (node *) a; 27 node *d = (node *) b; 28 int m = mul(point[0],*c,*d); 29 if(m == 0) return dis(*c,point[0]) - dis(*d,point[0]); 30 else return -m; 31 } 32 void gramham() 33 { 34 top = 1; 35 qsort(point + 1,n - 1,sizeof(node),cmp); 36 stack[0] = point[0]; 37 stack[1] = point[1]; 38 int i; 39 for(i = 2; i < n; i++) 40 { 41 while(top >= 1 && mul(stack[top - 1],stack[top],point[i]) <= 0) 42 top --; 43 stack[++ top] = point[i]; 44 } 45 } 46 int main() 47 { 48 int i,j; 49 int kemp; 50 int maxx; 51 //freopen("data.txt","r",stdin); 52 while(cin>>n) 53 { 54 memset(point,0,sizeof(point)); 55 memset(stack,0,sizeof(stack)); 56 kemp = 0; 57 for(i = 0; i < n; i++) 58 { 59 scanf("%d%d",&point[i].x,&point[i].y); 60 if(point[i].y < point[kemp].y || (point[i].y == point[kemp].y && point[i].x < point[kemp].x)) 61 kemp = i; 62 } 63 swap(point[0],point[kemp]); 64 gramham(); 65 maxx = 0; 66 for(i = 0; i <= top; i++) 67 { 68 for(j = i + 1; j <= top; j++) 69 maxx = max(maxx,dis(stack[i],stack[j])); 70 } 71 cout<<maxx<<endl; 72 } 73 return 0; 74 }