poj - 1113 - Wall

来源：互联网发布：日本对中国的援助数据编辑：程序博客网时间：2024/05/08 14:39

题意：顺时针方向给出N个点，求外围距离这些点L距离的点围成的图形的周长，结果四舍五入到整数(3 <= N <= 1000，1 <= L <= 1000，-10000 <= Xi, Yi <= 10000) 。

题目链接：http://poj.org/problem?id=1113

——>>先求凸包，然后求凸包的周长加上一个半径为L的圆的周长。

注意：用round()四舍五入后，若用%.0lf输出会WA，round()后强转为int后输出AC!

#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1000 + 10;const double eps = 1e-10;const double pi = acos(-1);int dcmp(double x){    if(fabs(x) < eps) return 0;    else return x < 0 ? -1 : 1;}struct Point{    double x;    double y;    Point(double x = 0, double y = 0):x(x), y(y){}    bool operator < (const Point& e) const{        return x < e.x || (dcmp(x - e.x) == 0 && y < e.y);    }}p[maxn], q[maxn];typedef Point Vector;Vector operator + (Point A, Point B){    return Vector(A.x + B.x, A.y + B.y);}Vector operator - (Point A, Point B){    return Vector(A.x - B.x, A.y - B.y);}Vector operator * (Point A, double p){    return Vector(A.x * p, A.y * p);}Vector operator / (Point A, double p){    return Vector(A.x / p, A.y / p);}double Cross(Vector A, Vector B){    return A.x * B.y - B.x * A.y;}int ConvexHull(Point *p, int n, Point* ch){        //求凸包    sort(p, p + n);    int 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--;    return m;}double Dis(Point A, Point B){    return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));}int main(){    int N, L;    while(scanf("%d%d", &N, &L) == 2){        for(int i = 0; i < N; i++) scanf("%lf%lf", &p[i].x, &p[i].y);        double ret = 0;        int m = ConvexHull(p, N, q);        for(int i = 1; i < m; i++) ret += Dis(q[i], q[i-1]);        ret += Dis(q[0], q[m-1]) + 2 * pi * L;        int ans = round(ret);        printf("%d\n", ans);    }    return 0;}