poj1113/hdu1348(凸包。。。两个网站上的输入输出有点出入)

Wall

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.

Sample Input

9 100
200 400
300 400
300 300
400 300
400 400
500 400
500 200
350 200
200 200

Sample Output

1628

Hint

结果四舍五入就可以了

这题是在求凸包的基础上，再求每个顶点对应角度的弧长，如图

外面围墙的直线部分与城墙的凸包长相同；
但是我们没有必要把每一个顶点对应的弧长都算出来。
注意因为每个顶对应的半径长都是l，所以一个多边形的所顶点对应弧长相加其实就是一个圆。
所以，最后总长为 凸包边长+π*2*l
最后需要的就是输出：

这里写代码片
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>

using namespace std;

const double eps=1e-8;
const double PI=acos(-1.0);
const int N=10100;
struct node{
    double x,y;
    node (double xx=0,double yy=0)
    {
        x=xx;y=yy;
    }
};
node po[N];
double l;
int n,top,sta[N*2];

node operator -(const node &a,const node &b)
{
    return node(a.x-b.x,a.y-b.y);
}

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

int cmp(const node &a,const node &b)
{
    if (dcmp(a.x-b.x)!=0) return a.x<b.x;
    else return a.y<b.y;
}

double Cross(node a,node b)
{
    return a.x*b.y-a.y*b.x;
}

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

void TuB()  //水平排序的凸包，实现起来简单
{
    top=0;
    memset(sta,0,sizeof(sta));
    int i;
    for (i=1;i<=n;i++)
    {
        while (top>1&&dcmp(Cross(po[i]-po[sta[top-1]],po[sta[top]]-po[sta[top-1]]))<=0) top--;
        sta[++top]=i;
    }
    int k=top;
    for (i=n-1;i>=1;i--)
    {
        while (top>k&&dcmp(Cross(po[i]-po[sta[top-1]],po[sta[top]]-po[sta[top-1]]))<=0) top--;
        sta[++top]=i; 
    }
    if (n>1) top--;  //n>1
    double ans=0;
    ans+=dis(po[sta[1]],po[sta[top]]);
    while (top>1) ans+=dis(po[sta[top]],po[sta[top-1]]),top--;
    ans+=(double)l*2*PI; //
    if(ans-(int)ans >= 0.5)
         printf("%d
",(int)ans+1);
    else
         printf("%d
",(int)ans);
    return;
}

int main()
{
    scanf("%d%lf",&n,&l);
    for (int i=1;i<=n;i++)
        scanf("%lf%lf",&po[i].x,&po[i].y);
    sort(po+1,po+1+n,cmp);
    TuB();
    return 0;
}