POJ 3348(计算几何+二维凸包+多边形面积)

问题描述:

Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are forced to save money on buying fence posts by using trees as fence posts wherever possible. Given the locations of some trees, you are to help farmers try to create the largest pasture that is possible. Not all the trees will need to be used.

However, because you will oversee the construction of the pasture yourself, all the farmers want to know is how many cows they can put in the pasture. It is well known that a cow needs at least 50 square metres of pasture to survive.

Input

The first line of input contains a single integer, n (1 ≤ n ≤ 10000), containing the number of trees that grow on the available land. The next n lines contain the integer coordinates of each tree given as two integers x and y separated by one space (where -1000 ≤ x, y ≤ 1000). The integer coordinates correlate exactly to distance in metres (e.g., the distance between coordinate (10; 11) and (11; 11) is one metre).

Output

You are to output a single integer value, the number of cows that can survive on the largest field you can construct using the available trees.

Sample Input

40 00 10175 075 101

Sample Output

151

题目题意:题目给我们n个点，问这个n个点围成的面积，能够喂养几头牛，一头牛需要50平方米空地。

题目分析:由题意构建二维凸包，然后求其面积。这个二维凸包的模板可以保存下。

代码如下:

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAXN=10000+200;struct point{    int x,y;};point list[MAXN];//凸包的点int stack[MAXN],top;//stack保存的是最后凸包的点的序号int cross(point p0,point p1,point p2) //计算叉积  p0p1 X p0p2{    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);}double dis(point p1,point p2)  //计算 p1p2的 距离{    return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));}bool cmp(point p1,point p2) //极角排序函数 ， 角度相同则距离小的在前面{    int tmp=cross(list[0],p1,p2);    if(tmp>0) return true;    else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true;    else return false;}void init(int n) //输入，并把  最左下方的点放在 list[0]  。并且进行极角排序{    int i,k;    point p0;    scanf("%d%d",&list[0].x,&list[0].y);    p0.x=list[0].x;    p0.y=list[0].y;    k=0;    for(i=1;i<n;i++)    {        scanf("%d%d",&list[i].x,&list[i].y);        if( (p0.y>list[i].y) || ((p0.y==list[i].y)&&(p0.x>list[i].x)) )        {            p0.x=list[i].x;            p0.y=list[i].y;            k=i;        }    }    list[k]=list[0];    list[0]=p0;    sort(list+1,list+n,cmp);}void graham(int n){    int i;    if(n==1) {top=0;stack[0]=0;}    if(n==2)    {        top=1;        stack[0]=0;        stack[1]=1;    }    if(n>2)    {        for(i=0;i<=1;i++) stack[i]=i;        top=1;        for(i=2;i<n;i++)        {            while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0) top--;            top++;            stack[top]=i;        }    }}int main(){    int n;    while (scanf("%d",&n)!=EOF) {        init(n);        graham(n);        stack[++top]=stack[0];        double area=0;        point o;        o.x=0,o.y=0;//多边形求面积        for (int i=0;i<top;i++) {            area+=cross(o,list[stack[i]],list[stack[i+1]]);        }        int ans=fabs(area)/100.0;//除以2，再除以50        printf("%d\n",ans);    }    return 0;}