[HDU 1348] Wall 凸包周长

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

题意：一个国王要求建筑师修一堵墙，围绕他的城堡，且墙与城堡之间的距离总不小于L输入节点和L求周长

思路：先求出凸包，然后求出周长加上一个半径为 L 的圆的周长，因为多边形只有再顶点处外面的墙才会形成圆弧，其他地方的墙和凸包的边平行。由于任意多边形外角和都是 360° 所以外面墙的周长就是凸包周长加上以 L 为半径的圆的周长。

[code]#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1010;

struct Point{
    int x, y;
    friend bool operator < (Point a, Point b){
        return a.y < b.y || (a.y == b.y && a.x < b.x);
    }
};

Point res[maxn];
Point point[maxn];

double mult(Point a, Point b, Point c){
    return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}

int graham(int n, Point *point, Point *res)
{
    sort(point, point+n);
    if(n == 0) return 0; res[0] = point[0];
    if(n == 1) return 1; res[1] = point[1];
    if(n == 2) return 2; res[2] = point[2];
    int top = 1;
    for(int i = 2; i < n; i++){
        while(top && mult(point[i], res[top], res[top-1]) >= 0){
            top--;
        }
        res[++top] = point[i];
    }
    int len = top;
    for(int i = n-2; i >= 0; i--){
        while(top > len && mult(point[i], res[top], res[top-1]) >= 0){
            top--;
        }
        res[++top] = point[i];
    }
    return top;
}

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

int main()
{
    int Test;
    cin>>Test;
    while(Test--){
        int n, l;
        cin>>n>>l;
        for(int i = 0; i < n; i++){
            cin>>point[i].x>>point[i].y;
        }
        int top = graham(n, point, res);
        double ans = 0;
        for(int i = 1; i <= top; i++){
            ans += length(res[i], res[i-1]);
        }
        ans += length(res[top], res[0]) + 2.0 * l * 3.1415926535897932384626;
        cout<<int(ans+0.5)<<endl;
        if(Test)
            cout<<endl;
    }
    return 0;
}