Nyoj cabling problem (minimum spanning tree)

。。。 is destined to do water problems.

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5#include <string>6#include <queue>7#include <algorithm>8#include <map>9#include <iomanip>Ten#include <climits> One#include <string.h> A#include <cmath> -#include <stdlib.h> -#include <vector> the#include <stack> -#include <Set> - using namespacestd; - #defineINF 1000000007 + #defineMAXN 4010 - #defineMod 1000007 + #defineN 100010 A #defineNN 30 at #defineSigma_size 3 - Const intMAX =1000100; - Const intMAXN = 6e5 +Ten; - using namespacestd; -typedefLong LongLL; -  in intvalue[555]; - intfa[555]; to structnode{ +     intu, V, c; -     BOOL operator< (ConstNode A)Const{ the         returnC <A.C; *     } $}a[255* -];Panax Notoginseng  -  the intT; + intN, M; A intans; the  + intFindset (intx) - { $     if(Fa[x] <0)returnx; $     returnFA[X] =Findset (fa[x]); - } - intMain () the { -CIN >>T;Wuyi      while(t--) { theAns =0; -Memset (A,0,sizeof(a)); Wumemset (Value,0,sizeof(value)); -memset (fa,-1,sizeof(FA)); AboutCIN >> N >>m; $          for(inti =0; I < m; ++i) { -scanf"%d%d%d", &a[i].u, &AMP;A[I].V, &a[i].c); -         } -          for(inti =1; I <= N; ++i) Ascanf"%d",&value[i]); +Sort (A, A +m); theSort (value +1, Value + N +1); -          for(inti =0; I < m; ++i) { $             intx =Findset (a[i].u); the             inty =Findset (A[I].V); the             if(X! =y) { theAns + =a[i].c; theFA[X] + =Fa[y]; -Fa[y] =x; in             } the         } theprintf"%d\n", ans+value[1]); About     } the     //System ("pause"); the     return 0; the}

Nyoj cabling problem (minimum spanning tree)