题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348

求凸包周长+2*PI*L;

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
const int N = ;
const double eps = 1e-;
const double PI = acos(-);
struct point
{
double x, y;
point(){}
point(double x, double y) : x(x), y(y) {}
point friend operator - (const point &p1, const point &p2)///矢量p2p1;
{
return point(p1.x-p2.x, p1.y-p2.y);
}
double friend operator ^ (const point &p1, const point &p2)///p1×p2;
{
return p1.x*p2.y - p1.y*p2.x;
}
}; point p[N], res[N]; double Dist(point p1, point p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
} int cmp1(point p1, point p2)///位置排序,找到最下方的;
{
if(p1.y != p2.y)
return p1.y < p2.y;
return p1.x < p2.x;///若有多个下方的找左边的;
}
int cmp2(point p1, point p2)///极角排序;若极角相同,距离近的在前面;
{
double k = (p1-p[])^(p2-p[]);
if( k>eps || (fabs(k)<eps && Dist(p1, p[]) < Dist(p2, p[]) ))
return ;
return ;
} int Graham(int n)///构造凸包,O(nlogn)
{
res[] = p[];if(n == ) return ;
res[] = p[];if(n == ) return ;
int top = ;
for(int i=; i<n; i++)
{
while(top && ((res[top]-res[top-])^(p[i]-res[top-])) <= ) top--;
///当res[top-1]->res[top]->p[i],出现右拐或直行时没说明res[top]不是凸包上的点;
res[++top] = p[i];
}
return top+;
} int main()
{
int n, T;
double L;
scanf("%d", &T);
while(T--)
{
scanf("%d%lf", &n, &L);
for(int i=; i<n; i++)
scanf("%lf %lf", &p[i].x, &p[i].y); sort(p, p+n, cmp1);///p[0]为最下方靠左的点;
sort(p+, p+n, cmp2);///以p[0]为基点,按叉积进行排序; int cnt = Graham(n);///求凸包的顶点个数cnt,保存在res中,下标从0开始; double ans = Dist(res[], res[cnt-]);
for(int i=; i<cnt; i++)
ans += Dist(res[i], res[i-]);
ans += PI**L;
printf("%.0f\n", ans);
if(T)puts("");
}
return ;
}
04-30 22:14