hdu 1392 Surround the Trees（ 凸包問題）

原題

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7987    Accepted Submission(s): 3044

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

 

代碼：

直接套用Graham算法模板即可，注意特殊考慮輸入點爲1和2的情況，wa了一次。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const double eps = 1e-8;
struct Point{
    double x, y;
};

const int MAXN = 200;
Point list[MAXN];
int stack[MAXN], top;


double crossProduct(Point a, Point b){
	return a.x*b.y - a.y*b.x;
}

int sgn(double x){
	if(fabs(x) < eps) return 0;
	if(x < 0) return -1;
	else return 1;
}

Point sub(Point a, Point b){
	Point p;
	p.x = a.x - b.x;
	p.y = a.y - b.y;
	return p;
}

double dist(Point a, Point b){
	return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

//相對於極點list[0]的極角排序
bool cmp(Point p1, Point p2){
    double temp  = crossProduct(sub(p1, list[0]), sub(p2, list[0]));
	if(sgn(temp)>0) return true;
	else if(sgn(temp)==0 && sgn(dist(p1, list[0])-dist(p2, list[0]))<=0) return true;
	else return false;
}

/*
* 求凸包，Graham算法
* 點的編號0~n-1
* 返回凸包結果Stack[0~top-1]爲凸包的編號
*/
void Graham(int n){
    Point p0 = list[0];
    int k = 0;
    for(int i=1;i<n;i++){
        if(p0.y>list[i].y || (p0.y==list[i].y && p0.x>list[i].x)){
            p0 = list[i];
            k = i;
        }
    }
    swap(list[k], list[0]);
    sort(list+1, list+n, cmp);

	stack[0] = 0;
    if(n==1){top = 1; return;}
    stack[1] = 1;
    if(n==2){top = 2; return;}

	top = 2;
	for(int i=2;i<n;i++){
        while(top>1 && sgn(crossProduct(sub(list[stack[top-1]], list[stack[top-2]]), sub(list[i], list[stack[top-2]])))<=0){
			top--;
        }
        stack[top++] = i;
	}
}

int main(){
	int n;
	while(1){
		scanf("%d", &n);
		if(n == 0) break;
        for(int i=0;i<n;i++){
            scanf("%lf %lf", &list[i].x, &list[i].y);
        }

		if(n==1){
			printf("0.00\n");
			continue;
		}
		if(n==2){
			printf("%.2lf\n", dist(list[0],list[1]));
			continue;
		}
        Graham(n);

        double ans = 0;
        for(int i=0;i<top-1;i++)
			ans += dist(list[stack[i]], list[stack[i+1]]);
		ans +=	dist(list[stack[0]], list[stack[top-1]]);
		printf("%.2lf\n", ans);
	}
	return 0;
}