【计算几何】Wall

DescriptionOnce upon a time there was a greedy King who ordered his chief Architect to build a wall aroundthe King's castle. The King was so greedy, that he would not listen to his Architect's proposalsto build a beautiful brick wall with a perfect shape and nice tall towers. Instead, he ordered tobuild the wall around the whole castle using the least amount of stone and labor, but demandedthat the wall should not come closer to the castle than a certain distance. If the King findsthat the Architect has used more resources to build the wall than it was absolutely necessaryto satisfy those requirements, then the Architect will loose his head. Moreover, he demandedArchitect to introduce at once a plan of the wall listing the exact amount of resources thatare needed to build the wall. 

Your task is to help poor Architect to save his head, by writing a program that will find theminimum possible length of the wall that he could build around the castle to satisfy King'srequirements.The task is somewhat simplified by the fact, that the King's castle has a polygonal shape andis situated on a flat ground. The Architect has already established a Cartesian coordinatesystem and has precisely measured the coordinates of all castle's vertices in feet.InputThe 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) isthe 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 containstwo integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that representthe coordinates of ith vertex. All vertices are different and the sides of the castle do notintersect anywhere except for vertices.OutputWrite to the output file the single number that represents the minimal possible length of thewall in feet that could be built around the castle to satisfy King's requirements. You mustpresent the integer number of feet to the King, because the floating numbers are not inventedyet. However, you must round the result in such a way, that it is accurate to 8 inches (1 footis equal to 12 inches), since the King will not tolerate larger error in the estimates.Sample Input9 100200 400300 400300 300400 300400 400500 400500 200350 200200 200Sample Output1628

这是一道凸包的入门题。
先将所有点按纵坐标从小到大排序，若纵坐标相同则按横坐标从小到大排序。

接着从头到尾扫描，保证每条边都在上一条边的逆时针方向（用栈来维护），得到凸包的右半集；再从尾到头扫描，同时保证每条边都在上一条边的逆时针方向，得到凸包的右半集部分。

由于这时栈顶多了一个首元素，所以舍弃栈顶。

Accode：

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <complex>using std::complex;const char fi[] = "wall.in";const char fo[] = "wall.out";const int maxN = 1010;const int MAX = 0x3f3f3f3f;const int MIN = ~MAX;const double zero = 1e-12;const double Pi = asin(1.L) * 2.L;typedef complex <double> Point;Point p[maxN];Point res[maxN];int n, R, top;void init_file(){    freopen(fi, "r", stdin);    freopen(fo, "w", stdout);    return;}void readdata(){    scanf("%d%d", &n, &R);    for (int i = 0; i < n; ++i)    {        double x, y;        scanf("%lf%lf", &x, &y);        p[i] = Point(x, y);    }    return;}template <typename _Tp>_Tp outer_product(const complex <_Tp> a,                  const complex <_Tp> b){    return a.real() * b.imag()    - a.imag() * b.real();}int cmp(const void *a, const void *b){    Point A = *(Point *)a,    B = *(Point *)b;    if (A.imag() < B.imag() - zero) return -1;    if (A.imag() > B.imag() + zero) return 1;    if (A.real() < B.real() - zero) return -1;    if (A.real() > B.real() + zero) return 1;    return 0;}void work(){    qsort(p, n, sizeof(p[0]), cmp);    res[0] = p[0];    res[1] = p[1];    top = 1;    for (int i = 2; i < n; ++i)    {        while (top &&               outer_product(p[i] - res[top],               res[top] - res[top - 1]) > -zero)            --top;        res[++top] = p[i];    }    int tmp = top;    res[++top] = p[n - 2];    for (int i = n - 3; i > -1; --i)    {        while (top > tmp &&               outer_product(p[i] - res[top],               res[top] - res[top - 1]) > -zero)            --top;        res[++top] = p[i];    }    double ans = Pi * (R << 1)    + abs(res[0] - res[top - 1]);    for (int i = 1; i < top; ++i)        ans += abs(res[i] - res[i - 1]);    printf("%d\n", (int)(ans + 0.5));    return;}int main(){    init_file();    readdata();    work();    return 0;}