POJ 3348 (凸包)

来源：互联网发布：后藤藤四郎极化数据编辑：程序博客网时间：2024/06/08 05:49
题意是求凸包的面积。
没什么坑，板子题
#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <cstdio>using namespace std;typedef unsigned long long ll;#define maxn 51111#define pi acos (-1)#define rotate Rotateconst double eps = 1e-8;int dcmp (double x) {    if (fabs (x) < eps)        return 0;    else return x < 0 ? -1 : 1;}struct point {    double x, y;    point (double _x = 0, double _y = 0) : x(_x), y(_y) {}    point operator - (point a) const {        return point (x-a.x, y-a.y);    }    point operator + (point a) const {        return point (x+a.x, y+a.y);    }    bool operator < (const point &a) const {        return x < a.x || (x == a.x && y < a.y);    }    bool operator == (const point &a) const {        return dcmp (x-a.x) == 0 && dcmp (y-a.y) == 0;    }};point operator * (point a, double p) {    return point (a.x*p, a.y*p);}double cross (point a, point b) {    return a.x*b.y-a.y*b.x;}double dot (point a, point b) {    return a.x*b.x + a.y*b.y;}int n, m, tot;point p[maxn], ch[maxn];int ConvexHull () {    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-1]) <= 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-1]) <= 0)            m--;        ch[m++] = p[i];    }    if (n > 1)        m--;    return m;}double dis (point a, point b) {    double xx = a.x-b.x, yy = a.y-b.y;    return sqrt (xx*xx + yy*yy);}double PolygonArea (point *p, int n) {    double ans = 0;    for (int i = 1; i < n-1; i++) {        ans += cross (p[i]-p[0], p[i+1]-p[0]);    }    return ans/2;}int main () {    //freopen ("in", "r", stdin);    ios::sync_with_stdio(0);    while (cin >> n) {        for (int i = 0; i < n; i++) {            cin >> p[i].x >> p[i].y;        }        m = ConvexHull ();        printf ("%lld\n", (long long)(floor)(PolygonArea (ch, m)/50.0));    }    return 0;}
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