zoukankan      html  css  js  c++  java
  • Wall(凸包POJ 1113)

    Wall
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 32360 Accepted: 10969

    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.

    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

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <stack>
    #include <algorithm>
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    
    const double Pi = 3.141592654;
    
    struct node
    {
        int x;
        int y;
    } P[500000],S[500000];
    
    int det(int x1,int y1,int x2,int y2)
    {
        return x1*y2-x2*y1;
    }
    
    int cross(node a,node b,node c)
    {
        return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y);
    }
    int dis(node a,node b)
    {
        return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
    }
    
    bool cmp(node a, node b)
    {
        int k=cross(P[0],a,b);
        if(k>0)
        {
            return true;
        }
        if(k==0&&dis(a,P[0])<dis(b,P[0]))
        {
            return true;
        }
        return false;
    }
    
    int main()
    {
        int n,m;
        int x,y,flag;
        while(~scanf("%d %d",&n,&m))
        {
            x=INF;
            y=INF;
            flag=0;
            for(int i=0; i<n; i++)
            {
                scanf("%d %d",&P[i].x,&P[i].y);
                if(y>P[i].y)
                {
                    x=P[i].x;
                    y=P[i].y;
                    flag=i;
                }
                else if(y==P[i].y&&x>P[i].x)
                {
                    x=P[i].x;
                    flag=i;
                }
            }
    
            node temp=P[flag];
            P[flag]=P[0];
            P[0]=temp;
            int top=0;
            if(n==2)
            {
                S[0]=P[0];
                S[1]=P[1];
                S[2]=S[0];
                top=2;
            }
            else if(n>2)
            {
                sort(P+1,P+n,cmp);
                P[n]=P[0];
                for(int i=0; i<2; i++)
                {
                    S[top++]=P[i];
                }
                top=1;
                for(int i=2; i<=n; i++)
                {
                    while(top>=1&&cross(S[top-1],S[top],P[i])<=0)
                    {
                        top--;
                    }
                    S[++top]=P[i];
                }
            }
            double MaxDis=0;
            for(int i=0; i<top; i++)
            {
                MaxDis+=sqrt(dis(S[i],S[i+1]));
            }
            MaxDis+=2*m*Pi;
            printf("%.0f
    ",MaxDis);
        }
        return 0;
    }
    
  • 相关阅读:
    C#中Excel的导入和导出的几种基本方式
    关于XML与类型Class的映射
    关于模板的相关注意事项
    OGRE 2.1 Windows 编译
    R6010 -abort() has been called
    VS2012 编译 Assimp
    C++ 并发消息队列
    Debian 安装下载工具软件
    Oracle SQL语句追踪
    记录自己学习网址
  • 原文地址:https://www.cnblogs.com/juechen/p/5255937.html
Copyright © 2011-2022 走看看