(hdu step 7.1.7)Wall(求凸包的周长——求将全部点围起来的最小凸多边形的周长)

Posted 2020-09-22 lxjshuju

题目：

Wall

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 119 Accepted Submission(s): 47

Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King‘s castle. The King was so greedy, that he would not listen to his Architect‘s proposals to build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall should not come closer to the castle than a certain distance. If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will loose his head. Moreover, he demanded Architect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall. Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King‘s requirements. The task is somewhat simplified by the fact, that the King‘s castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle‘s vertices in feet.

Input The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King‘s castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle. Next N lines describe coordinates of castle‘s vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.

Output Write to the output file the single number that represents the minimal possible length of the wall in feet that could be built around the castle to satisfy King‘s requirements. You must present the integer number of feet to the King, because the floating numbers are not invented yet. However, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates. This problem contains multiple test cases! The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks. The output format consists of N output blocks. There is a blank line between output blocks.

Sample Input 1 9 100 200 400 300 400 300 300 400 300 400 400 500 400 500 200 350 200 200 200

Sample Output 1628

Source Northeastern Europe 2001

Recommend JGShining

题目分析：

               求凸包的周长。再求图报的周长前，首先要做的是计算凸包——找到将全部点围起来的最小凸多边形。

对于找到凸包的算法，下面代码用的是graham算法，对这个算法不太熟悉的童鞋能够先看一下：

http://blog.csdn.net/hjd_love_zzt/article/details/44311333

代码例如以下：

/*
 * g.cpp
 *
 *  Created on: 2015年3月16日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const double epsi = 1e-8;
const double pi = acos(-1.0);
const int maxn = 1001;


struct PPoint{//结构体尽量不要定义成Point这样的,easy和C/C++本身中的变量同名
	double x;
	double y;

	PPoint(double _x = 0,double _y = 0):x(_x),y(_y){

	}

	PPoint operator - (const PPoint& op2) const{
		return PPoint(x - op2.x,y - op2.y);
	}

	double operator^(const PPoint &op2)const{
		return x*op2.y - y*op2.x;
	}
};


inline int sign(const double &x){
	if(x > epsi){
		return 1;
	}

	if(x < -epsi){
		return -1;
	}

	return 0;
}

inline double sqr(const double &x){
	return  x*x;
}


inline double mul(const PPoint& p0,const PPoint& p1,const PPoint& p2){
	return (p1 - p0)^(p2 - p0);
}


inline double dis2(const PPoint &p0,const PPoint &p1){
	return sqr(p0.x - p1.x) + sqr(p0.y - p1.y);
}

inline double dis(const PPoint& p0,const PPoint& p1){
	return sqrt(dis2(p0,p1));
}

int n;
double l;
PPoint p[maxn];
PPoint convex_hull_p0;


inline bool convex_hull_cmp(const PPoint& a,const PPoint& b){
	return sign(mul(convex_hull_p0,a,b)>0)|| sign(mul(convex_hull_p0,a,b)) == 0 && dis2(convex_hull_p0,a) < dis2(convex_hull_p0,b);
}


/**
 * 计算点集a[]的凸包b[]。当中点集a有n个元素
 */
int convex_hull(PPoint* a,int n,PPoint* b){
	if(n < 3){//假设顶点数小于3,构不成一个凸包
		//输出失败信息
		printf("wrong answer ,cause of n smaller than 3\n");
		return -1;
	}

	int i;
	for(i = 1 ; i < n ; ++i){//遍历点集中的每个点
		//寻找最低点(所谓的最低点就是最靠左下角的点)
		if(sign(a[i].x - a[0].x) < 0 || (sign(a[i].x - a[0].x) == 0 && sign(a[i].y < a[0].y) < 0 )){
			swap(a[i],a[0]);
		}
	}

	convex_hull_p0 = a[0];
	sort(a,a+n,convex_hull_cmp);//排序

	int newn = 2;
	b[0] = a[0];
	b[1] = a[1];

	/**
	 * 在剩下的点中不断前进,假设当前点在前进方向左側,
	 * 则将当前点进栈,否则将近期入栈的点出栈.知道当前点在前进方向的左側
	 */
	for(i = 2 ; i < n ; ++i){
		while(newn > 1 && sign(mul(b[newn-1],b[newn-2],a[i])) >= 0){
			newn--;
		}

		b[newn++] = a[i];//江当前点进栈
	}

	return newn;//返回栈顶指针
}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%d %lf",&n,&l);

		int i;
		for(i = 0 ; i < n ; ++i){
			scanf("%lf %lf",&p[i].x,&p[i].y);
		}

		n = convex_hull(p,n,p);
		p[n] = p[0];

		double ans = 0;
		for(i = 0 ; i < n ; ++i){//求凸包的周长
			ans += dis(p[i],p[i+1]);
		}


		ans += 2*pi*l;//加上外面围墙的周长

		/**
		 * "."后面的是小数精度控制。这里由于是浮点型。则取零代表不显示小数点（取整）
		 * .不为零时代表最大小数位数
		 */
		printf("%.0lf\n",ans);

		if(t != 0){//每个输出后面都要跟一个换行
			printf("\n");
		}
	}

	return 0;
}