func_code_string: #include<cstdio>   #include<vector>   #include<map>   #include<set>   #include<algorithm>   #include<cstdlib>   #include<cstring>   #include<cmath>   #include<string>   #include<deque>   #include<queue>   using namespace std;   #define db double   #define ll long long   int T;   struct node{       ll x,y;   }p[10];   int u[10], u2[10]; ll x[10],y[10],s[10], u3[10];   ll ans;      void c1(ll px, ll py) {       ll now = 0;       now += abs(px - x[u2[1]]) + abs(py - y[u2[1]]);       now += abs(px - x[u2[3]]);       now += abs(py - y[u2[2]]);       s[1] = u3[1] * (x[u2[2]] - px);       s[2] = u3[2] * (y[u2[3]] - py);       s[3] = u3[3] * (x[u2[4]] - px);       s[4] = u3[4] * (y[u2[4]] - py);       sort(s+1, s+5);       for(int i=1; i<=4; i++) now+=abs(s[i] - s[2]);       // if (now==4){       //     printf( %lld %lld n , px, py);       //     for(int i=1; i<=4; i++) printf( %d  ,u2[i]);       //     printf(  n );       //     for(int i=1; i<=4; i++) printf( %lld %lld n , x[i], y[i]);       //     printf(  n );                  // }          ans = min(ans, now);   }      int main() {       scanf( %d ,&T);       for(;T--;) {           int n = 4;           ans = 0;           for(int i=1; i<=n; i++) {               scanf( %lld %lld ,&p[i].x, &p[i].y);               ans += p[i].x + p[i].y;               u[i] = i;           }                      for(;1;) {               for(int i=1; i<=n; i++)                   x[i] = p[u[i]].x, y[i] = p[u[i]].y;               for(int i=1; i<=n; i++) u2[i] = i, u3[i] = 1;               c1(x[1], y[1]); c1(x[1], y[2]); c1(x[3], y[1]); c1(x[3], y[2]);                u2[1] = 2; u2[2] = 1; u2[3] = 4; u2[4] = 3;               u3[1] = u3[3] = -1;               c1(x[2], y[1]); c1(x[2], y[2]); c1(x[4], y[1]); c1(x[4], y[2]);                  u2[1] = 3; u2[2] = 4; u2[3] = 1; u2[4] = 2;               u3[1] = 1; u3[2] = -1; u3[3] = 1; u3[4] = -1;               c1(x[1], y[3]); c1(x[1], y[4]); c1(x[3], y[3]); c1(x[3], y[4]);               u2[1] = 4; u2[2] = 3; u2[3] = 2; u2[4] = 1;               u3[1] = u3[3] = -1;               u3[2] = u3[4] = -1;               c1(x[2], y[3]); c1(x[2], y[4]); c1(x[4], y[3]); c1(x[4], y[4]);               if (!next_permutation(u+1, u+1+n)) break;           }                      printf( %lld n , ans);       }       return 0;   }