POJ 1113&&HDU 1348

题意：凸包周长+一个完整的圆周长。因为走一圈，经过拐点时，所形成的扇形的内角和是360度，故一个完整的圆。

模板题，之前写的Graham模板不对，WR了很多发。。。。POJ上的AC代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define clc(a,b) memset(a,b,sizeof(a))
#define eps 1e-8
#define PI acos(-1.0)
typedef long long LL;
const int mod=47000;
const int inf=0x3f3f3f3f;
const int MAXN=1000;
using namespace std;

const int N = 1005;

int top;
struct point
{
    double x;
    double y;
} p[N], Stack[N];

double dis(point A, point B)
{
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}

double crossProd(point A, point B, point C)
{
    return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
}


int cmp(const void *a, const void *b)
{
    point *c = (point *)a;
    point *d = (point *)b;
    double k = crossProd(p[0], *c, *d);
    if (k<0 || !k && dis(p[0], *c)>dis(p[0], *d))
    return 1;
    return -1;
}

void Graham(int n)
{
    double x = p[0].x;
    double y = p[0].y;
    int mi = 0;
    for (int i=1; i<n; ++i)
    {
        if (p[i].x<x || (p[i].x==x && p[i].y<y))
        {
            x = p[i].x;
            y = p[i].y;
            mi = i;
        }
    }
    point tmp = p[mi];
    p[mi] = p[0];
    p[0] = tmp;
    qsort(p+1, n-1, sizeof(point), cmp);
    p[n] = p[0];
    Stack[0] = p[0];
    Stack[1] = p[1];
    Stack[2] = p[2];
    top = 2;
    for (int i=3; i<=n; ++i)
    {
        while (crossProd(Stack[top-1], Stack[top], p[i])<=0 && top>=2) --top;
        Stack[++top] = p[i];
    }
}

int main()
{
    int n, l;
    while (scanf("%d%d",  &n, &l) != EOF)
    {
            for (int i=0; i<n; ++i)
            {
                scanf ("%lf%lf", &p[i].x, &p[i].y);
            }
            Graham(n);
            double ans=0;
            for(int i=0;i<top;i++)
            {
                ans+=dis(Stack[i], Stack[i+1]);
            }
            ans += PI * (l + l);
            printf ("%d
", (int)(ans+0.5));
    }
    return 0;
}