POJ - 1113 Wall (凸包模板） Graham Scan 算法实现

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

简化下题意即求凸包的周长+2×PI×r。

这道题的答案是凸包周长加上一个圆周长，即包围凸包的一个圆角多边形，但是没弄明白那些圆角加起来为什么恰好是一个圆。每个圆角是以凸包对应的顶点为圆心，给定的L为半径，与相邻两条边的切点之间的一段圆弧。每个圆弧的两条半径夹角与对应的凸包的内角互补。假设凸包有n条边，则所有圆弧角之和为180°n-180°（n-2）=360°。故，围墙周长为=n条平行于凸包的线段+n条圆弧的长度=凸包周长+围墙离城堡距离L为半径的圆周长。

AC代码：学习自KB大佬的模板加以修改

// Author : RioTian
// Time : 20/10/21
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1000;
const double Pi = acos(-1.0);

struct point {
    int x, y;
    point() : x(), y() {}
    point(int x, int y) : x(x), y(y) {}
} list[N];
typedef point P;
int stack[N], top;
//计算叉积: p0p1 X p0p2
int cross(P p0, P p1, P p2) {
    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}
//计算 p1p2的 距离
double dis(P p1, P p2) {
    return sqrt((double)(p2.x - p1.x) * (p2.x - p1.x) +
                (p2.y - p1.y) * (p2.y - p1.y));
}
//利用极角排序，角度相同则距离小排前面
bool cmp(P p1, P p2) {
    int tmp = cross(list[0], p1, p2);
    if (tmp > 0)
        return true;
    else if (tmp == 0 && dis(list[0], p1) < dis(list[0], p2))
        return true;
    else
        return false;
}
//输入，并把  最左下方的点放在 list[0]  。并且进行极角排序
void init(int n) {
    int i, k = 0;
    cin >> list[0].x >> list[0].y;
    P p0 = list[0];  // p0 等价于 tmp 去寻找最左下方的点
    for (int i = 1; i < n; ++i) {
        cin >> list[i].x >> list[i].y;
        if (p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
            p0 = list[i], k = i;
    }
    list[k] = list[0];
    list[0] = p0;
    sort(list + 1, list + n, cmp);
}
//graham扫描法求凸包，凸包顶点存在stack栈中
//从栈底到栈顶一次是逆时针方向排列的
//如果要求凸包的一条边有2个以上的点
//那么要将while中的<=改成<
//但这不能将最后一条边上的多个点保留
//因为排序时将距离近的点排在前面
//那么最后一条边上的点仅有距离最远的会被保留，其余的会被出栈
//所以最后一条边需要特判
//如果要求逆凸包的话需要改cmp，graham中的符号即可
void Graham(int n) {
    int i;
    if (n == 1) top = 0, stack[0] = 0;
    if (n == 2) top = 1, stack[0] = 0, stack[1] = 1;
    if (n > 2) {
        for (i = 0; i <= 1; i++) stack[i] = i;
        top = 1;

        for (i = 2; i < n; i++) {
            while (top > 0 &&
                   cross(list[stack[top - 1]], list[stack[top]], list[i]) <= 0)
                top--;
            top++;
            stack[top] = i;
        }
    }
}
int main() {
    // freopen("in.txt", "r", stdin);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int N, L;
    while (scanf("%d%d", &N, &L) != EOF) {
        init(N);
        Graham(N);
        //叉积求凸包面积
        double res = 0;
        for (int i = 0; i < top; i++)
            res += dis(list[stack[i]], list[stack[i + 1]]);
        res += dis(list[stack[0]], list[stack[top]]);

        res += 2 * Pi * L;
        printf("%d\n", (int)(res + 0.5));
    }
}