POJ 2187 Beauty Contest【凸包周长】
2013-07-27 16:04
465 查看
题目:
http://poj.org/problem?id=1113http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/F
Wall
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 26219 | Accepted: 8738 |
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
题意:
按照顺时针顺序给你N个点的坐标,再给你一个长度LN个点代表城堡的坐标,
要求城堡任意一点到城墙的距离恰好 L 远建立城墙,求精确的长度
注意:结果四舍五入+0.5取整即可
思路:
凸包周长+以 L 为半径圆的周长/************************************************ Accepted 220 KB 0 ms C++ 1462 B 2013-07-27 15:46:32 题意:按照顺时针顺序给你N个点的坐标,再给你一个长度L N个点代表城堡的坐标, 要求城堡任意一点到城墙的距离恰好 L 远建立城墙,求精确的长度 注意:结果四舍五入+0.5取整即可 思路:凸包周长+以 L 为半径圆的周长 **********************************************/ #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; const int maxn = 1000+10; const double PI = 3.1415926535; int n,m; int L; struct Point{ double x,y; Point(){} Point(double _x, double _y) { x = _x; y = _y; } Point operator -(const Point &B) const { return Point(x-B.x, y-B.y); } }p[maxn], ch[maxn]; bool cmp(Point A, Point B) { if(A.x == B.x) return A.y < B.y; return A.x < B.x; } double dist(Point A, Point B) { return sqrt((A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y)); } double Cross(Point A, Point B) /**叉积*/ { return A.x*B.y - A.y*B.x; } void ConvexHull() /**求凸包*/ { sort(p,p+n,cmp); m = 0; for(int i = 0; i < n; i++) { while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; ch[m++] = p[i]; } int k = m; for(int i = n-2; i >= 0; i--) { while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; ch[m++] = p[i]; } if(n > 1) m--; } int solve() { ConvexHull(); double ans = 0; ch[m] = ch[0]; /**边界处理*/ for(int i = 0; i < m; i++) /**凸包周长*/ ans += dist(ch[i], ch[i+1]); ans += PI*L*2; /** 圆周长*/ return (int)(ans+0.5); /**四舍五入+0.5取整*/ } int main() { while(scanf("%d%d", &n,&L) != EOF) { for(int i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y); printf("%d\n", solve()); } return 0; }
相关文章推荐
- 【POJ 2187】Beauty Contest(凸包直径、旋转卡壳)
- poj2187 Beauty Contest 凸包求点集的直径
- poj 2187 Beauty Contest (凸包)
- POJ 2187 Beauty Contest 凸包+旋转卡壳
- POJ 2187 Beauty Contest(凸包直径)
- poj 2187 Beauty Contest 凸包 Graham
- POJ2187-Beauty Contest-凸包
- POJ 2187 Beauty Contest 凸包的应用
- poj 2187 Beauty Contest (凸包 Graham)
- POJ 2187 Beauty Contest (凸包)
- poj2187——Beauty Contest(凸包+旋转卡壳)
- poj 2187 Beauty Contest 题解(凸包模板+旋转卡壳)
- poj 2187 Beauty Contest(凸包)
- POJ 2187 Beauty Contest (凸包)
- poj 2187 Beauty Contest(凸包+旋转卡壳)
- poj 2187 Beauty Contest , 旋转卡壳求凸包的直径的平方
- POJ训练计划2187_Beauty Contest(几何/凸包)
- 【计算几何】 Andrew凸包算法 + 旋转卡壳(以求点对最长距离为例) -- 以 POJ 2187 Beauty Contest 为例
- poj 2187 Beauty Contest 经典题目:凸包+旋转卡壳
- poj 2187 Beauty Contest(二维凸包旋转卡壳)