POJ 3348 Cows 凸包面积
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POJ 3348 Cows 凸包面积
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題目大意:連接個點,求圍成的面積并除以50(向下取整)
題目思路:就是求凸包面積:按逆時針方向為凸包上每條邊指定方向,對于每條邊AB,累加(AXB)/2的值。
#include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<queue> #define INF 0x3f3f3f3f #define MAX 100005using namespace std;struct node {int x,y,id; }point[MAX];int n,ans[MAX],Stuck[MAX],top;bool cmp(struct node A,struct node B) {if(A.x < B.x)return true;else if(A.x==B.x && A.y < A.y)return true;return false; }int Cross(int x1,int y1,int x2,int y2,int x3,int y3) {return (x1-x2)*(y1-y3)-(x1-x3)*(y1-y2); }double Area() {double sum=0.0;for(int i=1;i<=top;i++)sum+=(point[Stuck[i-1]].x*point[Stuck[i]].y-point[Stuck[i]].x*point[Stuck[i-1]].y);return fabs(sum/2); }void Graham() {int len;top=1;for(int i=0;i<2;i++)Stuck[i]=i;for(int i=2;i<n;i++){while(top>0 && Cross(point[Stuck[top]].x,point[Stuck[top]].y,point[Stuck[top-1]].x,point[Stuck[top-1]].y,point[i].x,point[i].y)<=0)top--;Stuck[++top]=i;}len=top;Stuck[++top]=n-2;for(int i=n-3;i>=0;i--){while(top!=len && Cross(point[Stuck[top]].x,point[Stuck[top]].y,point[Stuck[top-1]].x,point[Stuck[top-1]].y,point[i].x,point[i].y)<=0)top--;Stuck[++top]=i;} }int main() {while(scanf("%d",&n)!=EOF){memset(ans,0,sizeof(ans));for(int i=0;i<n;i++){scanf("%d%d",&point[i].x,&point[i].y);point[i].id=i;}if(n<3){printf("0\n");continue;}sort(point,point+n,cmp);Graham();double ans=Area();int sum=ans/50;printf("%d\n",sum);}return 0; } View Code?
轉載于:https://www.cnblogs.com/alan-W/p/6021819.html
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