HDU 5839 Special Tetrahedron (2016CCPC online race 08) (Violence + pruning)

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5839

In a three-dimensional coordinate, give you n points, ask you how many tetrahedron (4 points, 6 edges) and satisfy at least four sides equal to the rest of the two nonadjacent.

The 4 cycle of violence, but in the 3rd cycle when the need to determine whether it is isosceles triangle, this is a pruning. In the 4th cycle of the time to determine whether the 4-point coplanar (cross-multiplication), 5 or 6 sides equal +1,4 sides equal to determine whether the other side of the intersection on the line.

After the game, I thought I was too much food.

1 //#pragma COMMENT (linker, "/stack:102400000, 102400000") 82#include <algorithm>3#include <iostream>4#include <cstdlib>5#include <cstring>6#include <cstdio>7#include <vector>8#include <cmath>9#include <ctime>Ten#include <list> One#include <Set> A#include <map> - using namespacestd; -typedefLong LongLL; thetypedef pair <int,int>P; - Const intN = 2e2 +5; - intCost[n][n];//The square of the side length of two points - structEdge { +     intx, y, Z; -}a[n*N]; +  A intMain () at { -     intT; -scanf"%d", &t); -      for(intCA =1; CA <= T; ++CA) { -         intN; -scanf"%d", &n); in          for(inti =1; I <= N; ++i) { -scanf" %d%d%d", &a[i].x, &AMP;A[I].Y, &a[i].z); to              for(intj =1; J < I; ++j) { +COST[I][J] = cost[j][i] = (a[i].x-a[j].x) * (a[i].x-a[j].x) + (A[I].Y-A[J].Y) * (A[I].Y-A[J].Y) + (a[i].z-a[j].z) * (A [I].z-a[j].z); -             } the         } * edge s1, S2, S3; $         intans, res =0, x1, x2, y1, y2, len1, len2;//X1 Y1 is a triangular face with different sides of the ends len1 for the isosceles triangle of the waistPanax Notoginseng          for(intI1 =1; I1 < n; ++I1) { -              for(intI2 = I1 +1; I2 < n; ++I2) { the                  for(inti3 = I2 +1; i3 < N; ++i3) { +                     //determine if 3 points form isosceles triangle A                     if(Cost[i1][i2]! = Cost[i1][i3] && cost[i2][i3]! = Cost[i1][i2] && cost[i2][i3]! =Cost[i1][i3]) the                         Continue; +                     Else if(Cost[i1][i2] = = Cost[i1][i3] && Cost[i1][i2] = =Cost[i2][i3]) { -Len1 = Cost[i1][i2], x1 = y1 =0; $                     } $                     Else if(Cost[i1][i2] = =Cost[i2][i3]) { -Len1 = Cost[i1][i2], x1 = i1, y1 =i3; -                     } the                     Else if(Cost[i1][i2] = =Cost[i1][i3]) { -Len1 = Cost[i1][i2], x1 = i2, y1 =i3;Wuyi                     } the                     Else { -Len1 = Cost[i2][i3], x1 = i1, y1 =I2; Wu                     } -                      for(intI4 = i3 +1; I4 <= N; ++I4) { About                         if(COST[I1][I4]! = Cost[i2][i4] && cost[i1][i4]! = Cost[i3][i4] && cost[i3][i4]! =Cost[i2][i4]) $                             Continue; -                         Else if(Cost[i1][i4] = = Cost[i2][i4] && Cost[i1][i4] = =Cost[i3][i4]) { -Len2 = Cost[i1][i4], x2 = y2 =0; -                         } A                         Else if(Cost[i1][i4] = =Cost[i2][i4]) { +Len2 = Cost[i1][i4], x2 = i3, y2 =I4; the                         } -                         Else if(Cost[i1][i4] = =Cost[i3][i4]) { $Len2 = Cost[i1][i4], x2 = i2, y2 =I4; the                         } the                         Else { theLen2 = Cost[i2][i4], x2 = i1, y2 =I4; the                         } -s1.x=a[i2].x-a[i1].x;s1.y=a[i2].y-a[i1].y;s1.z=a[i2].z-a[i1].z;  ins2.x=a[i3].x-a[i1].x;s2.y=a[i3].y-a[i1].y;s2.z=a[i3].z-a[i1].z;  thes3.x=a[i4].x-a[i1].x;s3.y=a[i4].y-a[i1].y;s3.z=a[i4].z-a[i1].z;  theans=s1.x*s2.y*s3.z+s1.y*s2.z*s3.x+s1.z*s2.x*s3.y-s1.z*s2.y*s3.x-s1.x*s2.z*s3.y-s1.y*s2.x*s3.z;  About                         if(ans = =0)//whether 4 points are coplanar the                             Continue; the                         if(!x1 | |!x2) {//6 sides or 5 sides equal the                             if(Len1 = =len2) +res++; -                         } the                         Else {Bayi                             if(Len1 = = Len2 && (x1! = x2 && X1! = y2 && y1! = x2 && y1! = y2))//Four edges equal theres++; the                         } -                     } -                 } the             } the         } theprintf"Case #%d:%d\n", CA, res); the     } -     return 0; the}

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HDU 5839 Special Tetrahedron (2016CCPC online race 08) (Violence + pruning)