Optimal irrigation for CCF simulation problems

Optimal irrigation

Time limit:	1.0s
Memory Limit:	256.0MB

Problem DescriptionRay Ray contracted a lot of wheat fields, in order to irrigate the wheat fields, Ray Ray dug a very deep well in the first wheat field, all the wheat fields from this well to water diversion irrigation.
In order to irrigate, Ray Ray need to establish some canals to connect wells and wheat fields, and Ray Ray can also use some of the wheat fields as "transit stations", using canals to connect different fields of wheat, so as long as a wheat field can be irrigated, the wheat fields connected to it can also be irrigated.
Now Ray Ray know which fields of wheat can be built between canals and the cost of building each channel (note that not all wheat fields can be established between canals). How much is it necessary to irrigate all the wheat fields to build canals? Input FormatThe first line of input contains two positive integers n, m, respectively, representing the number of slices in the wheat field and the amount of canals that Ray Ray can establish. Wheat fields use 1, 2, 3, ... Label in turn.
Next m line, each line contains three integers ai, bi, CI, which indicates that a channel can be established between the AI wheat field and the second Bi wheat field, the cost of which is CI. output FormatThe output line contains an integer that represents the minimum cost required to irrigate all wheat fields. Sample Input4 4
1 2 1
2 3 4
2 4 2
3 4 3 Sample Output6 Sample DescriptionThe following three canals were established: wheat field 1 with wheat field 2, wheat field 2 with wheat field 4, wheat field 4 with wheat Field 3. Evaluation use case size and the first 20% evaluation use cases meet: n≤5.
The first 40% evaluation cases meet: N≤20.
The first 60% evaluation cases meet: n≤100.
All evaluation cases meet: 1≤n≤1000,1≤m≤100,000,1≤ci≤10,000. Problem solving: minimum spanning tree

1#include <bits/stdc++.h>2 #defineLL Long Long3 #defineINF 0x3f3f3f3f3f3f3f4 #definePII pair<int,int>5 using namespacestd;6 Const intMAXN = -;7 structarc{8     intTo,cost,next;9Arcintx =0,inty =0,intz =-1){Tento =x; OneCost =y; ANext =Z; -     } -}e[500000]; the intTOT,N,M,HEAD[MAXN]; - LL D[MAXN]; - BOOLDONE[MAXN]; - voidAddintUintVintCost ) { +E[tot] =arc (V,cost,head[u]); -Head[u] = tot++; +E[tot] =arc (U,cost,head[v]); AHEAD[V] = tot++; at } -priority_queue< pii,vector< PII >,greater< PII > >Q; - LL Prim () { -      while(!q.empty ()) Q.pop (); -      for(inti =1; I <= N; ++i) { -Done[i] =false; inD[i] =INF; -     } toQ.push (Make_pair (d[1] =0,1)); +LL ans =0; -      while(!Q.empty ()) { the         intU =Q.top (). Second; * Q.pop (); $         if(Done[u])Continue;Panax NotoginsengDone[u] =true; -Ans + =D[u]; the          for(inti = Head[u]; ~i; i =E[i].next) { +             if(!done[e[i].to] && E[i].cost <D[e[i].to]) { AD[e[i].to] =E[i].cost; the Q.push (Make_pair (d[e[i].to],e[i].to)); +             } -         } $     } $     returnans; - } - intMain () { the     intu,v,w; -      while(~SCANF ("%d%d",&n,&m)) {Wuyimemset (head,-1,sizeof(head)); the          for(inti = tot =0; I < m; ++i) { -scanf" %d%d%d",&u,&v,&W); Wu Add (u,v,w); -         } Aboutprintf"%i64d\n", Prim ()); $     } -     return 0; -}

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Optimal irrigation for CCF simulation problems