Random Point in Triangle
題目描述
Bobo has a triangle ABC with A(x1,y1),B(x2,y2)A(x1,y1),B(x2,y2) and C(x3,y3)C(x3,y3). Picking a point P uniformly in triangle ABC, he wants to know the expectation value E=max{SPAB,SPBC,SPCA}E=max{SPAB,SPBC,SPCA} where SXYZSXYZ denotes the area of triangle XYZ.
Print the value of 36×E36×E. It can be proved that it is always an integer.
輸入描述:
The input consists of several test cases and is terminated by end-of-file.
Each test case contains six integers x1,y1,x2,y2,x3,y3x1,y1,x2,y2,x3,y3.
* |x1|,|y1|,|x2|,|y2|,|x3|,|y3|≤108|x1|,|y1|,|x2|,|y2|,|x3|,|y3|≤108
* There are at most 105105 test cases.
輸出描述:
For each test case, print an integer which denotes the result.
輸入
0 0 1 1 2 2
0 0 0 0 1 1
0 0 0 0 0 0
輸出
0
0
0
題目描述:
多組輸入三角形各個頂點座標p1,p2,p3,在三角形中任取一點p,計算 期望E=max(S(p,p1,p2),max(S(p,p1,p3),S(p,p2,p3)));
思路:
1、產生隨機點,選出在三角形內部的點,對每個符合條件的點分別求出最大面積並求和,最後取面積和的平均值即爲期望E。 (當隨機數足夠多時,所有最大面積和的平均值即爲期望)
2、根據下面找規律代碼可以求出:(由於是跑的隨機數,結果定不完全等於22)
3、求出E後可以看出規律,即 36*E=22*S(p1,p2,p3)
補充:
1、根據三角形座標求三角形面積:向量差乘求三角形面積
2、座標點結構體:
struct poLL{
LL x;
LL y;
};
struct poLL p1,p2,p3,p;
p1.x=x1,p1.y=y1;
p2.x=x2,p2.y=y2;
p3.x=x3,p3.y=y3;
一看就能看懂,挺好用的,在傳點座標時不用分別傳x、y了
找規律代碼:
#include<bits/stdc++.h>
using namespace std;
struct point{
double x;
double y;
};
double S(point p1,point p2,point p3) ///向量叉乘求三角形面積可以看上面鏈接
{
return fabs((p1.x-p3.x)*(p1.y-p2.y)-(p1.x-p2.x)*(p1.y-p3.y));
}
bool judge(point p,point p1,point p2,point p3) ///判斷生成的隨機點是不是在三角中 原理是:三個
///小三角形面積之和是不是等於原來大的三角形面積
{
if(S(p1,p2,p3)==(S(p1,p2,p)+S(p,p1,p3)+S(p2,p3,p))){
//printf("*%lf %lf %lf %lf\n",S(p1,p2,p3),S(p1,p2,p),S(p,p1,p3),S(p2,p3,p));
return true;
}
return false;
}
int main()
{
double x1,x2,x3,y1,y2,y3;
while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF){
double cnt=0;
struct point p1,p2,p3,p;
p1.x=x1,p1.y=y1;
p2.x=x2,p2.y=y2;
p3.x=x3,p3.y=y3;
srand(time(0));
double sum=0;
for(int i=0;i<100000000;i++){
int x=rand();
int y=rand();
p.x=x*1.0,p.y=y*1.0;
if(judge(p,p1,p2,p3)){
cnt++;
sum+=(max(S(p1,p2,p),max(S(p,p1,p3),S(p2,p3,p))));
}
//printf("%.3lf\n",sum);
}
double S1=S(p1,p2,p3);
double q=((sum/cnt)*36.0)/S1;
printf("%.3lf\n",q);
}
return 0;
}
AC代碼:
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
struct poLL{
LL x;
LL y;
};
LL S(poLL p1,poLL p2,poLL p3)
{
return abs((p1.x-p3.x)*(p1.y-p2.y)-(p1.x-p2.x)*(p1.y-p3.y));
}
int main()
{
LL x1,x2,x3,y1,y2,y3;
while(scanf("%lld%lld%lld%lld%lld%lld",&x1,&y1,&x2,&y2,&x3,&y3)!=EOF){
struct poLL p1,p2,p3,p;
p1.x=x1,p1.y=y1;
p2.x=x2,p2.y=y2;
p3.x=x3,p3.y=y3;
printf("%lld\n",11*S(p1,p2,p3));
}
return 0;
}