Collision Detection
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1207 Accepted Submission(s): 367
Problem Description
In physical simulations, video games and computational geometry, collision detection involves algorithms for checking for collision, i.e. intersection, of two given objects. Collision detection algorithms are a basic component of 3D video games. Without them, characters could go through walls and other obstacles.
Here comes an interesting problem, given a ball and a cuboid, you need to detect whether they collide. We say that two objects collide if and only if they share at least one point.
Input
The first line of input is the number of test cases.
Each test case contains two lines, the first line contains 24 integers X1, Y1, Z1, …, X8, Y8, Z8, representing the 8 vertexes of the cuboid. Vertexes are given in random order and you can make sure that all edges of the cuboid are parallel to coordinate axes; the second line contains 4 integers X,Y,Z,R representing the central point of the ball and its radius. All integers given are non-negative and will be less than 100000.
Output
For each case output "Yes" Or "No" on a single line.
Sample Input
2
0 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 1
2 2 2 2
0 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 1
2 2 2 1
Sample Output
Yes
No
Source
2008 Asia Chengdu Regional Contest Online
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真是日了狗了。。。
读错题wa到死。。。
英文差的原因？
我以为长方体是空心的。。
然后球是有可能在内部
我说我怎么写的那么麻烦，又是判断点，又是判断边，又是判断面。。。。
然后全推了重写。。。
又WA到死。。。
最后发现又是强制转化类型的问题。。。
如果变量是int 类型，即使一出赋值给long long ，在赋值之前的计算也会溢出。。。
所以不溢出的办法就是之前的int 就写成 long long 类型的。。。
1 /*************************************************************************
2 > File Name: code/2015summer/#6/II.cpp
3 > Author: 111qqz
4 > Email: [email protected]
5 > Created Time: 2015年08月01日 星期六 03时09分18秒
6 ************************************************************************/
7
8 #include<iostream>
9 #include<iomanip>
10 #include<cstdio>
11 #include<algorithm>
12 #include<cmath>
13 #include<cstring>
14 #include<string>
15 #include<map>
16 #include<set>
17 #include<queue>
18 #include<vector>
19 #include<stack>
20 #define y0 abc111qqz
21 #define y1 hust111qqz
22 #define yn hez111qqz
23 #define j1 cute111qqz
24 #define tm crazy111qqz
25 #define lr dying111qqz
26 using namespace std;
27 #define REP(i, n) for (int i=0;i<int(n);++i)
28 typedef long long LL;
29 typedef unsigned long long ULL;
30 const int inf = 0x7fffffff;
31
32
33 LL dis(LL x1,LL x2,LL y1,LL y2,LL z1,LL z2)
34 {
35
36 return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)+(z1-z2)*(z1-z2);
37
38 }
39 int main()
40 {
41 int T; //整体算法是寻找距离球最近的点，比较这个点到球心的距离和半径
42 // 。。。判断了边+点+面还没A的我好傻逼啊。。。。。。。。sad
43 // 卧槽，读错题了，长方体是实心的啊喂！怪不得我想的那么麻烦！
44 cin>>T;
45 while (T--)
46 {
47 LL tx,ty,tz;
48 LL x1 = inf;
49 LL x2 = -inf;
50 LL y1 = inf;
51 LL y2 = -inf;
52 LL z1 = inf;
53 LL z2 = -inf;
54 for ( int i = 1 ; i <= 8 ; i++ )
55 {
56
57 scanf("%lld %lld %lld",&tx,&ty,&tz);
58 x1 = min(tx,x1);
59 x2 = max(tx,x2);
60 y1 = min(ty,y1);
61 y2 = max(ty,y2);
62 z1 = min(tz,z1);
63 z2 = max(tz,z2);
64 }
65 LL bx,by,bz,r;
66 LL x,y,z;
67 scanf("%lld %lld %lld %lld",&bx,&by,&bz,&r);
68 x = bx;
69 y = by;
70 z = bz;
71 if (x>x2) x=x2;
72 if (x<x1) x=x1;
73 if (y>y2) y=y2;
74 if (y<y1) y=y1;
75 if (z>z2) z=z2;
76 if (z<z1) z=z1;
77 if (r*r>=dis(x,bx,y,by,z,bz))
78 {
79 puts("Yes");
80 }
81 else
82 {
83 puts("No");
84 }
85
86
87 }
88
89 return 0;
90 }