poj1113

题意：给出平面上若干个点的坐标，让建一个环形围墙，把所有的点围在里面，且距所有点的距离不小于l。求围墙的最小长度。

分析：凸包周长＋半径为l的圆周长

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
usingnamespace std;

#define maxn 1005

struct point
{
    double x, y;
} pnt[maxn], res[maxn];

int n, l;

bool mult(point sp, point ep, point op)
{
    return (sp.x - op.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - op.y);
}

booloperator<(const point &l, const point &r)
{
    return l.y < r.y || (l.y == r.y && l.x < r.x);
}

int graham(point pnt[], int n, point res[])
{
    int i, len, top =1;
    sort(pnt, pnt + n);
    if (n ==0)
        return0;
    res[0] = pnt[0];
    if (n ==1)
        return1;
    res[1] = pnt[1];
    if (n ==2)
        return2;
    res[2] = pnt[2];
    for (i =2; i < n; i++)
    {
        while (top && mult(pnt[i], res[top], res[top -1]))
            top--;
        res[++top] = pnt[i];
    }
    len = top;
    res[++top] = pnt[n -2];
    for (i = n -3; i >=0; i--)
    {
        while (top != len && mult(pnt[i], res[top], res[top -1]))
            top--;
        res[++top] = pnt[i];
    }
    return top;
}

double dis(point &p, point &q)
{
    double x1 = p.x - q.x, y1 = p.y - q.y;
    return sqrt(x1 * x1 + y1 * y1);
}
int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d%d", &n, &l);
    for (int i =0; i < n; i++)
        scanf("%lf%lf", &pnt[i].x, &pnt[i].y);
    int tot = graham(pnt, n, res);
    double ans = l *2*3.14159265;
    for (int i =0; i < tot; i++)
        ans += dis(res[i], res[(i +1) % tot]);
    printf("%.0f\n", ans);
    return0;
}