zoukankan      html  css  js  c++  java
  • POJ

    Description

    Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered to build the wall around the whole castle using the least amount of stone and labor, but demanded that the wall should not come closer to the castle than a certain distance. If the King finds that the Architect has used more resources to build the wall than it was absolutely necessary to satisfy those requirements, then the Architect will loose his head. Moreover, he demanded Architect to introduce at once a plan of the wall listing the exact amount of resources that are needed to build the wall.
    img
    Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King's requirements.

    The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet.

    Input

    The first line of the input file contains two integer numbers N and L separated by a space. N (3 <= N <= 1000) is the number of vertices in the King's castle, and L (1 <= L <= 1000) is the minimal number of feet that King allows for the wall to come close to the castle.

    Next N lines describe coordinates of castle's vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.

    Output

    Write to the output file the single number that represents the minimal possible length of the wall in feet that could be built around the castle to satisfy King's requirements. You must present the integer number of feet to the King, because the floating numbers are not invented yet. However, you must round the result in such a way, that it is accurate to 8 inches (1 foot is equal to 12 inches), since the King will not tolerate larger error in the estimates.

    Sample Input

    9 100
    200 400
    300 400
    300 300
    400 300
    400 400
    500 400
    500 200
    350 200
    200 200
    

    Sample Output

    1628
    

    Hint

    结果四舍五入就可以了

    Source

    Northeastern Europe 2001

    简化下题意即求凸包的周长+2×PI×r。

    这道题的答案是凸包周长加上一个圆周长,即包围凸包的一个圆角多边形,但是没弄明白那些圆角加起来为什么恰好是一个圆。每个圆角是以凸包对应的顶点为圆心,给定的L为半径,与相邻两条边的切点之间的一段圆弧。每个圆弧的两条半径夹角与对应的凸包的内角互补。假设凸包有n条边,则所有圆弧角之和为180°n-180°(n-2)=360°。故,围墙周长为=n条平行于凸包的线段+n条圆弧的长度=凸包周长+围墙离城堡距离L为半径的圆周长。

    AC代码:学习自KB大佬的模板加以修改

    // Author : RioTian
    // Time : 20/10/21
    #include <algorithm>
    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <iostream>
    using namespace std;
    const int N = 1000;
    const double Pi = acos(-1.0);
    
    struct point {
        int x, y;
        point() : x(), y() {}
        point(int x, int y) : x(x), y(y) {}
    } list[N];
    typedef point P;
    int stack[N], top;
    //计算叉积: p0p1 X p0p2
    int cross(P p0, P p1, P p2) {
        return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
    }
    //计算 p1p2的 距离
    double dis(P p1, P p2) {
        return sqrt((double)(p2.x - p1.x) * (p2.x - p1.x) +
                    (p2.y - p1.y) * (p2.y - p1.y));
    }
    //利用极角排序,角度相同则距离小排前面
    bool cmp(P p1, P p2) {
        int tmp = cross(list[0], p1, p2);
        if (tmp > 0)
            return true;
        else if (tmp == 0 && dis(list[0], p1) < dis(list[0], p2))
            return true;
        else
            return false;
    }
    //输入,并把  最左下方的点放在 list[0]  。并且进行极角排序
    void init(int n) {
        int i, k = 0;
        cin >> list[0].x >> list[0].y;
        P p0 = list[0];  // p0 等价于 tmp 去寻找最左下方的点
        for (int i = 1; i < n; ++i) {
            cin >> list[i].x >> list[i].y;
            if (p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
                p0 = list[i], k = i;
        }
        list[k] = list[0];
        list[0] = p0;
        sort(list + 1, list + n, cmp);
    }
    //graham扫描法求凸包,凸包顶点存在stack栈中
    //从栈底到栈顶一次是逆时针方向排列的
    //如果要求凸包的一条边有2个以上的点
    //那么要将while中的<=改成<
    //但这不能将最后一条边上的多个点保留
    //因为排序时将距离近的点排在前面
    //那么最后一条边上的点仅有距离最远的会被保留,其余的会被出栈
    //所以最后一条边需要特判
    //如果要求逆凸包的话需要改cmp,graham中的符号即可
    void Graham(int n) {
        int i;
        if (n == 1) top = 0, stack[0] = 0;
        if (n == 2) top = 1, stack[0] = 0, stack[1] = 1;
        if (n > 2) {
            for (i = 0; i <= 1; i++) stack[i] = i;
            top = 1;
    
            for (i = 2; i < n; i++) {
                while (top > 0 &&
                       cross(list[stack[top - 1]], list[stack[top]], list[i]) <= 0)
                    top--;
                top++;
                stack[top] = i;
            }
        }
    }
    int main() {
        // freopen("in.txt", "r", stdin);
        // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
        int N, L;
        while (scanf("%d%d", &N, &L) != EOF) {
            init(N);
            Graham(N);
            //叉积求凸包面积
            double res = 0;
            for (int i = 0; i < top; i++)
                res += dis(list[stack[i]], list[stack[i + 1]]);
            res += dis(list[stack[0]], list[stack[top]]);
    
            res += 2 * Pi * L;
            printf("%d
    ", (int)(res + 0.5));
        }
    }
    

    The desire of his soul is the prophecy of his fate
    你灵魂的欲望,是你命运的先知。

  • 相关阅读:
    C#获取配置文件中的文件数据
    wpf MVVMLight的DataGrid绑定数据
    扫码支付自动跳转,可以使用第三方网站实现扫码二维码付款然后跳转到想要的页面展示想要内容或者是解压码或者是某个资源的下载页呢 具体步骤(我以你上传一个压缩包到某种网盘或者可以下载的地址等让人付费解压为例):
    oracle数据库如何创建用户以及分配权限
    ORA-12547: TNS: 丢失连接
    springmvc中applicationapplicationContext头部代码
    No mapping found for HTTP request with URI
    在Navicat新建用户
    myeclipse 项目引入 com.sun.image.codec.jpeg 的api报错解决方法
    java.lang.NullPointerException at org.apache.jsp.**_jsp.jspInit(**_jsp.java)tomcat启动异常解决方法
  • 原文地址:https://www.cnblogs.com/RioTian/p/13854465.html
Copyright © 2011-2022 走看看