【poj3348】 Cows

http://poj.org/problem?id=3348 (题目链接)

题意

　　给出平面上n个点，以这n个点中的一些围成的多边形面积 div 50的最大值。

Solution

　　凸包求面积。

　　很好做，构造完凸包后从栈底开始向上求叉乘之和，也就是将凸包分成许多小三角形求面积和。

代码

// poj3348
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<map>
#define inf 2147483640
#define LL long long
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
    LL x=0,f=1;char ch=getchar();
    while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=10010;
struct point {int x,y;}p[maxn];
int s[maxn],n,top;

int cross(point p0,point a,point b) {
    return (a.x-p0.x)*(b.y-p0.y)-(a.y-p0.y)*(b.x-p0.x);
}
double dis(point a,point b) {
    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp(point a,point b) {
    int t=cross(p[1],a,b);
    if (t>0) return 1;
    else if (t==1 && dis(a,p[1])<dis(b,p[1])) return 1;
    else return 0;
}
void Graham() {
    if (n==1) s[top=1]=1;
    else if (n==2) {s[top=1]=1;s[++top]=2;}
    else {
        s[top=1]=1;s[++top]=2;
        for (int i=3;i<=n;i++) {
            while (top>1 && cross(p[s[top-1]],p[s[top]],p[i])<=0) top--;
            s[++top]=i;
        }
    }
}
double Size() {
    point p0=p[s[1]];
    double sum=0;
    for (int i=3;i<=top;i++) sum+=cross(p0,p[s[i-1]],p[s[i]]);
    return sum/2;
}
int main() {
    while (scanf("%d",&n)!=EOF) {
        int k=1;
        for (int i=1;i<=n;i++) {
            scanf("%d%d",&p[i].x,&p[i].y);
            if (p[i].x<p[k].x || (p[i].x==p[k].x && p[i].y<p[k].y)) k=i;
        }
        point p0=p[k];
        p[k]=p[1];p[1]=p0;
        sort(p+2,p+1+n,cmp);
        Graham();
        printf("%d\n",(int)Size()/50);
    }
    return 0;
}