专题 计算几何学 凸包 hdu1392Surround the Trees

题目

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1392
Problem Description
There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him?
The diameter and length of the trees are omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line.

There are no more than 100 trees.

Input
The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer is less than 32767. Each pair is separated by blank.

Zero at line for number of trees terminates the input for your program.

Output
The minimal length of the rope. The precision should be 10^-2.

Sample Input
9
12 7
24 9
30 5
41 9
80 7
50 87
22 9
45 1
50 7
0

Sample Output
243.06

思路

• 简单凸包题，按着模板来就行
• 注意特殊情况，及只有一个点和两个点的部分

代码

#include<iostream>#include<cstdio>#include<cstring>//#include<vector>#include<algorithm>#include<cmath>#define U_DEBUG#define L_JUDGE#ifdef L_JUDGE#pragma warning(disable:4996)#endifusing namespace std;const int MAXN=2e5+10;class Node{    public:        int x,y;        Node(int a=0,int b=0){            x=a;y=b;        }        Node(const Node &p){            x=p.x;y=p.y;        }};Node pt[MAXN];int N;int stack[MAXN];int top=0;bool IsEmpty(){    return top==0;}void Push(int x){    top++;    stack[top]=x;}int NextToTop(){    return stack[top-1];}int Top(){    return stack[top];}void Pop(){    top--;}// true: not exchange;  false: exchangebool Cmp(const Node &p1,const Node &p2){    int ret=(p2.x-pt[0].x)*(p1.y-pt[0].y)        -(p2.y-pt[0].y)*(p1.x-pt[0].x);    return ret<0;}int Direction(const Node &p0,const Node &p1,const Node &p2){    return (p2.x-p0.x)*(p1.y-p0.y)        -(p2.y-p0.y)*(p1.x-p0.x);}double Distance(const Node &a,const Node &b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main(){    #ifdef L_JUDGE        freopen("in.txt","r",stdin);        //freopen("out.txt","w",stdout);    #endif        while(scanf("%d",&N),N){            int mini=0;            for(int i=0;i<N;i++){                scanf("%d%d",&pt[i].x,&pt[i].y);                if(pt[i].y<pt[mini].y){                    mini=i;                }            }            if(N==1){                printf("%.2lf\n",0.0);                continue;            }else if(N==2){                printf("%.2lf\n",Distance(pt[0],pt[1]));                continue;            }            swap(pt[0],pt[mini]);            sort(pt+1,pt+N,Cmp);/*            for(int i=1;i<N;i++){                for(int j=i+1;j<N;j++){                    printf("%d\n",Direction(pt[0],pt[i],pt[j]));                }            }*/            top=0;            Push(0);Push(1);Push(2);            int p1,p2;            for(int i=3;i<N;i++){                p2=Top();                p1=NextToTop();                while(Direction(pt[p1],pt[p2],pt[i])>=0){                    Pop();                    p2=Top();                    p1=NextToTop();                }                Push(i);            }            int tmpx=Top();            int a,b;            a=Top();Pop();            double ans=0;            while(!IsEmpty()){                b=Top();Pop();                ans+=Distance(pt[a],pt[b]);                a=b;            }            ans+=Distance(pt[tmpx],pt[a]);            printf("%.2lf\n",ans);        }    #ifdef L_JUDGE        fclose(stdin);        fclose(stdout);        system("out.txt");    #endif    return 0;}