Lightoj1239

题目链接：点这里
题意：给你一些点，把这些点全部包住，并且离点的距离为d，求周长。
思路：和poj1113 这道题很想。求出凸包后结果在加上一个半径为d的圆的周长。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#include <set>
#define ll long long
#define llu unsigned long long
using namespace std;
const int maxn = 100010;
const double eps = 1e-8;
const double PI  = acos(-1.0);
struct Point {
    double x,y;
    Point (double x = 0,double y = 0) : x(x),y(y) {}
};
Point p[60050],s[60500];
struct Line{
    Point a,b;
};
Line line[100];
int indl,indp;

bool dcmp(double x) {
    if(fabs(x)-0.0 <= eps) return true;
    return false;
}

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

double cross(Point a,Point b,Point c) {
    if(dcmp((b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y))) return 0;
    return ((b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y));
}
bool cmp1(Point a,Point b) {
    if(a.x != b.x) return a.x < b.x;
    else return a.y<b.y;
}
bool cmp2(Point a,Point b) {
    double m = cross(p[1],a,b);
    if(dcmp(m)) {
        return dis(p[1],b) > dis(p[1],a) ? true : false;
    }else {
        if(m > 0) return true;
        else return false;
    }
}
int n;
int main(){
    int T; scanf("%d",&T);
    for(int cas = 1; cas<=T; cas++){
        scanf("%d",&n);
        double d;
        scanf("%lf",&d);
        for(int i=1;i<=n;i++) {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        if(n == 1) {
            printf("Case %d: %.10lf\n",cas,(PI*2*d)); continue;
        }
        sort(p+1,p+1+n,cmp1);
        sort(p+2,p+n+1,cmp2);
        s[1] = p[1];
        s[2] = p[2];
        int top = 2;
        for(int i=3;i<=n;i++) {
            while(top>=2 && cross(s[top-1],s[top],p[i]) <= 0) top--;
            s[++top] = p[i];
        }
        double l = 0;
        for(int i=2;i<=top;i++) {
            l += dis(s[i],s[i-1]);
        }
//        printf("%lf\n",l);
        l += dis(s[1],s[top]);
        l += (PI*2*d);
        printf("Case %d: %.10lf\n",cas,l);
    }
    return 0;
}


/*
2
1 2
0 0
2 2
0 0
1 1
*/