[HDU 1348] Wall 凸包周长

来源：互联网发布：js currenttimemillis 编辑：程序博客网时间：2024/04/30 03:55

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。

#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;}

0 0