Solid Geometry Homework

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5298

Description

Yellowstar is studying solid geometry recently,and today’s homework is about the space,plane and sphere.So he draw many planes and spheres in the draft paper.These infinite planes and (the surface of)spheres divides the whole drawing space(which can be considered as a infinite 3D-space) into many disjoint regions.Planes and spheres forms the borders of these regions,and they don’t belong to any regions.
Then he comes up with a crazy idea:color the whole space with crayons.He wants that one region has only one color,and two adjacent regions should be colored differently (“adjacent” means the area of two regions’ common borders is greater than zero).Unfortunately,he has only two crayons:a yellow one and a red one.
Yellowstar likes yellow very much,so he gives some coordinates.The regions these points belong to should be colored yellow.
Given positions of all the planes and spheres and the coordinates mentioned above.You should determine:Is there a way to satisfy all the requests?Yellowstar also gives some other coordinates.He wants to know which color they will be while all the requests are satisfied.

Input

The first line contains an integer T,denoting the number of the test cases.
For each test case, the first line contains 4 integers m,n,p and q, denoting the number of planes,spheres,points and queries.
Then m lines follows,each containing four integers a,b,c and d,denoting the linear equation(ax+by+cz+d=0) of this plane.|a|+|b|+|c|>0.
Then n lines follows,each containing four integers x,y,z and r,denoting the center coordinate(x,y,z) and radius of this sphere.
Then p lines follows, each containing three integers x,y,z,denoting point(x,y,z),the region it belongs to should be colored yellow.
Next q lines are queries.Each contains three integers x,y,z-the coordinate of this point.You need to output which color it will be.

T<=30,0<=m<=100,0<=n<=10,0<=p<=200,1<=q<=2000,|all given numbers|<=10^6，any two planes or spheres aren’t coincidence.No point lies on given planes or spheres.
There is a blank line before each case.

Output

For each case，if there is no such a coloring way to color the whole space and meet all the requests,print“Impossible”.
Otherwise，for each query，print a line.If the color of this point can be certainly inferred,print it(’Y’ for yellow or ’R’ for red)；if not(both are possible),print”Both”.
Print a blank line between adjacent cases.

Sample Input

3

1 1 1 2
0 0 1 0
0 0 0 2
0 0 1
0 0 -1
0 0 4

1 1 2 1
0 0 1 0
0 0 0 2
0 0 1
0 0 -1
0 0 4

1 1 0 2
0 0 1 0
0 0 0 2
0 0 4
0 0 -1

Sample Output

R
R

Impossible

Both
Both

Hint

题意

在一个三维平面上有一堆平面，有一堆圆，然后这些玩意儿把平面切成了很多块。

然后每一块要么是红色，要么是黄色。

相邻的两块颜色不同。

现在已知p个点的颜色是黄色。

然后问你接下来q个点的颜色是啥。

题解：

首先其实这个空间的颜色分布已经被那p个点唯一确认了。

所以我们只要知道一个区域的颜色就好了。

因为只有两种颜色，判断一个点的颜色只要知道和这些圆的位置关系和这些平面的位置关系就好了。

然后这道题就结束了……

大概就是这样 喵。

代码

#include<bits/stdc++.h>
using namespace std;

struct node
{
    long long a,b,c,d;
}plane[120],circle[12],P[2005],P2[205];

int n,m,p,q;

int check_plane(node A,node B)
{
    return A.a*B.a+A.b*B.b+A.c*B.c+A.d>0?1:0;
}

int check_cirle(node A,node B)
{
    return (A.a-B.a)*(A.a-B.a)+(A.b-B.b)*(A.b-B.b)+(A.c-B.c)*(A.c-B.c)>A.d*A.d?1:0;
}

int check(node a)
{
    int ans = 0;
    for(int i=0;i<m;i++)
        ans^=check_plane(plane[i],a);
    for(int i=0;i<n;i++)
        ans^=check_cirle(circle[i],a);
    return ans;
}
void solve()
{
    scanf("%d%d%d%d",&m,&n,&p,&q);
    for(int i=0;i<m;i++)
        scanf("%lld%lld%lld%lld",&plane[i].a,&plane[i].b,&plane[i].c,&plane[i].d);
    for(int i=0;i<n;i++)
        scanf("%lld%lld%lld%lld",&circle[i].a,&circle[i].b,&circle[i].c,&circle[i].d);
    if(p==0){
        for(int i=0;i<q;i++)
        {
            scanf("%lld%lld%lld",&P[i].a,&P[i].b,&P[i].c);
            printf("Both\n");
        }
        return;
    }
    for(int i=0;i<p;i++)
        scanf("%lld%lld%lld",&P2[i].a,&P2[i].b,&P2[i].c);
    for(int i=0;i<q;i++)
        scanf("%lld%lld%lld",&P[i].a,&P[i].b,&P[i].c);
    int flag = check(P2[0]);
    for(int i=0;i<p;i++)
        if(check(P2[i])^flag==1)
        {
            printf("Impossible\n");
            return;
        }
    for(int i=0;i<q;i++)
    {
        if(check(P[i])^flag==1)
            printf("R\n");
        else
            printf("Y\n");
    }
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        solve();
        if(t)puts("");
    }
    return 0;
}