第一行一个整数n，代表岛屿数量。
接下来n-1行，每行三个整数u,v,w，代表u号岛屿和v号岛屿由一条代价为c的桥梁直接相连，保证1<=u,v<=n且1<=c<=100000。
第n+1行，一个整数m，代表敌方机器能使用的次数。
接下来m行，每行一个整数ki，代表第i次后，有ki个岛屿资源丰富，接下来k个整数h1,h2,…hk，表示资源丰富岛屿的编号。
输出有m行，分别代表每次任务的最小代价。
对于100%的数据，2<=n<=250000,m>=1,sigma(ki)<=500000,1<=ki<=n-1
Stage2 day2
这个应该是虚树的入门题吧
1 program repair(input,output); 2 const 3 inf=123456789; 4 type 5 etype=record 6 t,next:longint; 7 w:int64; 8 end; 9 var 10 b,e:array[0..500050]of etype; 11 father:array[0..500050,0..20]of longint; 12 v:array[0..250025]of boolean; 13 a,d,head,q,dfn,fa:array[0..250025]of longint; 14 w,f:array[0..250025]of int64; 15 c:array[0..500050]of longint; 16 n,m,i,j,k,x,y,cnt,top,h,t,r:longint; 17 z:int64; 18 procedure add(x,y:longint;w:int64); 19 begin 20 inc(cnt);e[cnt].t:=y;e[cnt].w:=w;e[cnt].next:=a[x];a[x]:=cnt; 21 end; 22 function min(a,b:int64):int64; 23 begin 24 if a<b then exit(a) else exit(b); 25 end; 26 procedure dfs(k:longint); 27 var 28 i:longint; 29 begin 30 inc(cnt);dfn[k]:=cnt;if d[k]>r then r:=d[k]; 31 i:=a[k]; 32 while i<>0 do 33 begin 34 if dfn[e[i].t]=0 then 35 begin 36 father[e[i].t,0]:=k;d[e[i].t]:=d[k]+1;w[e[i].t]:=min(w[k],e[i].w); 37 dfs(e[i].t); 38 end; 39 i:=e[i].next; 40 end; 41 end; 42 procedure sort(q,h:longint); 43 var 44 i,j,x,t:longint; 45 begin 46 i:=q;j:=h;x:=dfn[c[(i+j)>>1]]; 47 repeat 48 while dfn[c[i]]<x do inc(i); 49 while x<dfn[c[j]] do dec(j); 50 if i<=j then 51 begin 52 t:=c[i];c[i]:=c[j];c[j]:=t; 53 inc(i);dec(j); 54 end; 55 until i>j; 56 if j>q then sort(q,j); 57 if i<h then sort(i,h); 58 end; 59 function lca(x,y:longint):longint; 60 var 61 t:longint; 62 begin 63 if d[x]<d[y] then begin t:=x;x:=y;y:=t; end; 64 for k:=r downto 0 do if d[father[x,k]]>=d[y] then x:=father[x,k]; 65 if x=y then exit(x); 66 for k:=r downto 0 do if father[x,k]<>father[y,k] then begin x:=father[x,k];y:=father[y,k]; end; 67 exit(father[x,0]); 68 end; 69 procedure ins(x,y:longint); 70 begin 71 inc(cnt);b[cnt].t:=y;b[cnt].next:=head[x];head[x]:=cnt; 72 end; 73 begin 74 assign(input,‘repair.in‘);assign(output,‘repair.out‘);reset(input);rewrite(output); 75 readln(n);cnt:=0; 76 fillchar(a,sizeof(a),0); 77 for i:=2 to n do begin read(x,y,z); add(x,y,z);add(y,x,z); end; 78 fillchar(dfn,sizeof(dfn),0);cnt:=0;father[1,0]:=0;d[1]:=1; 79 for i:=1 to n do w[i]:=inf;r:=0; 80 dfs(1); 81 k:=0;i:=1; 82 while i<r do begin i:=i+i;inc(k); end; 83 r:=k; 84 for j:=1 to r do for i:=1 to n do father[i,j]:=father[father[i,j-1],j-1]; 85 readln(m); 86 fillchar(head,sizeof(head),0);fillchar(v,sizeof(v),false);fillchar(f,sizeof(f),0); 87 for i:=1 to m do 88 begin 89 read(x);for j:=1 to x do begin read(c[j]);v[c[j]]:=true; end; 90 sort(1,x);y:=x; 91 for j:=2 to x do begin inc(y);c[y]:=lca(c[j],c[j-1]);if c[y]=1 then dec(y); end; 92 sort(1,y);x:=1; 93 for j:=2 to y do if c[j]<>c[j-1] then begin inc(x);c[x]:=c[j]; end; 94 top:=1;q[1]:=1;cnt:=0; 95 for j:=1 to x do 96 begin 97 while lca(q[top],c[j])<>q[top] do dec(top); 98 fa[c[j]]:=q[top];ins(q[top],c[j]); 99 inc(top);q[top]:=c[j]; 100 end; 101 h:=0;t:=1;q[1]:=1; 102 while h<t do 103 begin 104 inc(h); 105 j:=head[q[h]]; 106 while j<>0 do begin inc(t);q[t]:=b[j].t;j:=b[j].next; end; 107 end; 108 for j:=x+1 downto 2 do 109 begin 110 if (w[q[j]]<f[q[j]]) or v[q[j]] then f[q[j]]:=w[q[j]]; 111 inc(f[fa[q[j]]],f[q[j]]); 112 end; 113 writeln(f[1]); 114 for j:=1 to x do begin v[c[j]]:=false;f[c[j]]:=0;head[c[j]]:=0; end;f[1]:=0;head[1]:=0; 115 end; 116 close(input);close(output); 117 end.
原文：http://www.cnblogs.com/Currier/p/6614028.html