bzoj2753: [SCOI2012] ski with time capsule

bfs+ Minimum tree graph +kruskal algorithm.

The smallest tree diagram is an image of the smallest generation of graphs, this can not take Kruskal algorithm or prim algorithm directly, otherwise it will be wrong.

Is the case of w[u][v]!=w[v][u].

And this problem with the Zhu Liu algorithm is certainly not feasible.

But the direction of the problem is not the nature of the side, but the height of the point is determined. This allows us to layer the minimum spanning tree.

If you add a point with a height of H, simply use the Kruskal algorithm to select the shortest edge, and it will not affect the subsequent selection.

So we changed the Kruskal algorithm to the end point height for the first keyword descending and the edge length for the second keyword ascending sort ... (You know, or CMP, um.) )

So it can be spicy.

#include <cstdio>#include<algorithm>#include<cstring>using namespacestd;Const intMAXN =100000+Ten;Const intMAXM =2000000+Ten;intRead () {intx=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}structEdge {intu,v,d; }E[MAXM];intH[MAXN],F[MAXN];intG[maxn],v[maxm],next[maxm],eid;intres1,n,m;Long LongRes2;BOOLVIS[MAXN];intQ[maxm],l,r,u;voidAddedge (intAintb) {V[eid]=b; Next[eid]=g[a]; g[a]=eid++;}BOOLCMP (Edge A,edge b) {if(H[A.V]!=H[B.V])returnH[a.v]>H[B.V]; returna.d<B.D; }intFindintx) {returnf[x]==x?x:f[x]=find (F[x]);} intMain () {memset (g,-1,sizeof(g)); N=read (); m=read ();  for(intI=1; i<=n;i++) h[i]=read ();  for(intI=1, u,v;i<=m;i++) {u=read (); V=read (); E[i].d=read (); if(h[u]>=H[v]) Addedge (U,V); if(h[v]>=H[u])        {Addedge (v,u); swap (u,v);} E[I].U=u; e[i].v=v; } sort (E+1, e+m+1, CMP); Vis[q[r++]=1]=1, res1=1;  while(l<r) {u=q[l++];  for(intI=g[u];~i;i=next[i])if(!Vis[v[i]]) {Vis[q[r++]=v[i]]=1; Res1++; }    }         for(intI=1; i<=n;i++) f[i]=i;  for(intI=1, ru,rv;i<=m;i++)if(vis[e[i].u]&&VIS[E[I].V]) {ru=find (E[I].U); RV=find (E[I].V); if(ru!=RV) {Res2+=E[I].D; F[RV]=Ru; }} printf ("%d%lld\n", Res1,res2); return 0;}

bzoj2753: [SCOI2012] ski with time capsule