POJ 1113 WALL

 

Wall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13254		Accepted: 4217

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

 

http://acm.pku.edu.cn/JudgeOnline/problem?id=1113

 

/* 计算几何，求凸包 这题的结果等于这个多边形构成的凸包的周长加上以所给半径为半径的圆的周长 步骤如下： 1)算法首先寻找最最靠下方的点，如果遇到y坐标相同，则寻找x坐标最小的点firstP 2)然后根据所有点相对于firstP的偏角的大小进行排序，遇到偏角相等的，只取距离 firstP最远的点(排序利用自己手写的快排) 3)然后利用Graham算法求凸包 4)最后直接求职 */ #include <iostream> #include <cmath> #define PI 3.1415926 #define MAX_N 1000 using namespace std; //存储原始输入的坐标值，rad是输入的半径 double cord[MAX_N + 2][2], rad; int seq[MAX_N + 2]; int stack[MAX_N + 2]; int n, top; int firstP; int realN; void swap(int pos1, int pos2) { int temp = seq[pos1]; seq[pos1] = seq[pos2]; seq[pos2] = temp; } int dir(int nodes, int node1, int node2) { double x1 = cord[node1][0], y1 = cord[node1][1]; double x2 = cord[node2][0], y2 = cord[node2][1]; double sx = cord[nodes][0], sy = cord[nodes][1]; return (x2 - sx) * (y1 - sy) - (x1 - sx) * (y2 - sy); } double getDist(int node1, int node2) { double x1 = cord[node1][0], y1 = cord[node1][1]; double x2 = cord[node2][0], y2 = cord[node2][1]; double res = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); return res; } int compare(int node1, int node2) { double x1 = cord[node1][0], y1 = cord[node1][1]; double x2 = cord[node2][0], y2 = cord[node2][1]; double sx = cord[firstP][0], sy = cord[firstP][1]; double type = dir(firstP, node1, node2); if(type == 0) { double dist1 = (x1 - sx) * (x1 - sx) + (y1 - sy) * (y1 - sy); double dist2 = (x2 - sx) * (x2 - sx) + (y2 - sy) * (y2 - sy); if(dist1 > dist2) return -2; else if(dist1 == dist2) return 0; else return 2; } else if(type > 0) return 1; else return -1; } void fastSort(int start, int end) { if(start < end) { int curPos = start; int posS = start, posE = end + 1; while(true) { while(compare(seq[++posS], seq[curPos]) < 0 && posS < end); while(compare(seq[--posE], seq[curPos]) > 0 && posE > start); if(posS < posE) swap(posS, posE); else break; } swap(curPos, posE); fastSort(start, posE - 1); fastSort(posE + 1, end); } } void sortSeq() { int i, s = 0; for(i = 1; i <= n; i++) { //最低最左点不参加排序 if(i == firstP) continue; seq[++s] = i; } realN = n - 1; fastSort(1, realN); //清理夹角相同但是距离不同的点，只取举例firstP最远的点 i = 1; while(i < realN) { s = i + 1; //equal angle but smaller distance while(s <= realN && compare(seq[i], seq[s]) == -2) { seq[s] = -1; //置为无效 s++; } i = s; } } //寻找凸包 void findQ() { int nodes, node1, node2, type; top = 0; stack[top++] = firstP; int s = 1; int c = 0; while(c < 2) { if(seq[s] != -1) { c++; stack[top++] = seq[s]; } s++; } for(; s <= realN; s++) { if(seq[s] == -1) continue; while(true) { nodes = stack[top - 2]; node1 = stack[top - 1]; node2 = seq[s]; type = dir(nodes, node1, node2); if(type >= 0) top--; else break; } stack[top++] = seq[s]; } } double getRes() { double totalDist = 0; int lastNode = firstP; int curNode; while(top > 0) { curNode = stack[--top]; totalDist += getDist(lastNode, curNode); lastNode = curNode; } //totalDist += getDist(lastNode, firstP); totalDist += 2 * PI * rad; return totalDist; } int main() { int i; cin>>n>>rad; int minX = INT_MAX, minY = INT_MAX; for(i = 1; i <= n; i++) { cin>>cord[i][0]>>cord[i][1]; if((cord[i][1] < minY) || (cord[i][1] == minY && cord[i][0] < minX)) { firstP = i; minX = cord[i][0]; minY = cord[i][1]; } } sortSeq(); findQ(); double res = getRes(); printf("%.0f/n", res); return 0; }