2544: Length of Rope_please input the length of the rope: 12m-优快云博客

本文介绍了一个经典的几何问题——如何计算一根绳子的长度，该绳子围绕着一个凸多边形的所有顶点（被视为半径为R的钉子），并紧密贴合这些钉子。文章给出了输入数据格式，包括顶点数量、钉子半径及位置，并提供了输出要求及示例。

2544: Length of Rope

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	32768K	174	58	Standard

There is a convex polygon whose vertices are all nails. The nails are of radius R. We need a rope to round these nails tightly. You can see an example here.

Input

The input contains several cases. Each case start with an integer N and a real number R. N is the number of vertices of the polygon. R is the radius of nails. 1<=N<=200. Each of the next N lines is two real numbers, standing for the position of the corresponding nail. The nails are listed clockwise or counterclockwise.

Output

The length of the rope with two digits precision (after a decimal point). Each answer takes a line.

Sample Input

4 1
0.0 0.0
2.0 0.0
2.0 2.0
0.0 2.0

Sample Output

14.28

Problem Source: liuctic

#include<stdio.h>
#include<math.h>
double dis(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
const double pi=3.1415926;
int main()
{
    double x[210],y[210];
    double r;
    int n,i;
    while(scanf("%d%lf",&n,&r)==2)
    {
        double l=pi*r*2;
        //printf("%.2lf/n",l);
        scanf("%lf%lf",&x[1],&y[1]);
        for(i=2;i<=n;i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
            l+=dis(x[i-1],y[i-1],x[i],y[i]);
            //printf("%.2lf/n",l);
        }
        //scanf("%lf%lf",&x[n],&y[n]);
        l+=dis(x[1],y[1],x[n],y[n]);
        printf("%.2lf/n",l);
    }
    return 0;
}