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【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;}

  

【poj3348】 Cows