poj3384

题意：求凸包面积/50，并取整。

分析：用模板。

View Code

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

using namespace std;



#define maxn 10005



struct Point

{

    double x, y;

} point[maxn], cvx[maxn];



int n, m;



bool mult(Point sp, Point ep, Point op)

{

    return (sp.x - ep.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - ep.y);

}



bool operator <(const Point &l, const Point &r)

{

    return l.y < r.y || (l.y == r.y && l.x < r.x);

}



int graham(Point pnt[], int n, Point res[])

{

    int i, len, top = 1;

    sort(pnt, pnt + n);

    if (n == 0)

        return 0;

    res[0] = pnt[0];

    if (n == 1)

        return 1;

    res[1] = pnt[1];

    if (n == 2)

        return 2;

    res[2] = pnt[2];

    for (i = 2; i < n; i++)

    {

        while (top && mult(pnt[i], res[top], res[top - 1]))

            top--;

        res[++top] = pnt[i];

    }

    len = top;

    res[++top] = pnt[n - 2];

    for (i = n - 3; i >= 0; i--)

    {

        while (top != len && mult(pnt[i], res[top], res[top - 1]))

            top--;

        res[++top] = pnt[i];

    }

    return top;

    // 返回凸包中点的个数

}



void input()

{

    scanf("%d", &n);

    for (int i = 0; i < n; i++)

        scanf("%lf %lf", &point[i].x, &point[i].y);

}



double areaofp(int vcount, Point plg[])

{

    int i;

    double s;

    if (vcount < 3)

        return 0;

    s = plg[0].y * (plg[vcount - 1].x - plg[1].x);

    for (i = 1; i < vcount; i++)

        s += plg[i].y * (plg[(i - 1)].x - plg[(i + 1) % vcount].x);

    return s / 2;

}



int main()

{

    //freopen("t.txt", "r", stdin);

    input();

    m = graham(point, n, cvx);

    printf("%d\n", (int)(areaofp(m, cvx) / 50));

    return 0;

}