Wall
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
题目类型：求凸包
算法分析：城堡的围墙长度等于城堡凸包的周长再加上半径为L的圆的周长。这是因为国王的城堡是一个凸多边形，非转角处的围墙可以平行于城堡的边，总长度为凸包周长。而每个转角处的围墙长度是其所占弧的周长，城堡转角处的角度之和为2Pi。所以所有转角处的弧周长之和为圆的周长
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/************************************************* Author :supermaker Created Time :2016/1/26 15:38:22 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1000 + 66; int sgn (double val) { if (fabs (val) < EPS) return 0; else if (val < 0) return -1; return 1; } struct Point { double x, y; Point () {} Point (double xx, double yy) : x (xx), y (yy) {} Point operator + (const Point &a) const { return Point (x + a.x, y + a.y); } Point operator - (const Point &a) const { return Point (x - a.x, y - a.y); } double operator ^ (const Point &a) const { return x * a.y - y * a.x; } double operator * (const Point &a) const { return x * a.x + y * a.y; } }; double Dist (Point p1, Point p2) { return sqrt ((p2 - p1) * (p2 - p1)); } Point p[maxn]; int n, L; stack <int> sta; bool polarcmp (Point p1, Point p2) { double tt = (p1 - p[1]) ^ (p2 - p[1]); if (sgn (tt) == 1) return true; else if (sgn (tt) == 0 && sgn (Dist (p1, p[1]) - Dist (p2, p[1])) <= 0) return true; return false; } void Garham () { Point start = p[1]; int start_pos = 1; for (int i = 1; i <= n; i++) if (sgn (p[i].x - start.x) == -1 || (sgn (p[i].x - start.x) == 0 && sgn (p[i].y - start.y) == -1)) start = p[i], start_pos = i; swap (p[start_pos], p[1]); sort (p + 2, p + 1 + n, polarcmp); while (!sta.empty ()) sta.pop (); sta.push (1), sta.push (2); for (int i = 3; i <= n; i++) { while (sta.size () >= 2) { int tt1 = sta.top (); sta.pop (); int tt2 = sta.top (); sta.pop (); if (sgn ((p[tt1] - p[tt2]) ^ (p[i] - p[tt2])) <= 0) sta.push (tt2); else { sta.push (tt2), sta.push (tt1); break; } } sta.push (i); } } int main() { //CFF; //CPPFF; while (scanf ("%d%d", &n, &L) != EOF) { for (int i = 1; i <= n; i++) { double x, y; scanf ("%lf%lf", &x, &y); p[i] = Point (x, y); } Garham (); double res = 2 * PI * L; int per = sta.top (); sta.pop (); int start = per, cur; while (!sta.empty ()) { cur = sta.top (); sta.pop (); res += Dist (p[per], p[cur]); per = cur; } res += Dist (p[start], p[cur]); printf ("%.lf\n", res); } return 0; } |
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