POJ 2187 Beauty Contest【凸包周长】
题目:
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 26219 | Accepted: 8738 |
Description
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 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
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
Source
题意:
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:
/************************************************
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;
}
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