[Codevs 1036] Business travel

"Problem description"

A businessman in a capital city often goes to the towns to do business, and they do it on their own lines to save time.

Suppose there are N towns, the capital is numbered 1, the merchant departs from the capital, there are road connections between the other towns, and if there are direct links between any two towns, it takes a unit of time to travel between them. The country has a well-developed road network, which can reach any town from the capital, and the road network will not be a ring.

Your task is to help the businessman calculate his shortest travel time.

Input

the first line in the input file has an integer n,1<=n<=30 000, which is the number of towns.

N-1 lines below, each line consists of two integers a and b (1<=a, b<=n; a<>b), indicating that town A and town B have road connections.

In the N+1 Act an integer m, the following M-line, each line has the number of towns that the merchant needs to pass sequentially.

Output

outputs the shortest time the merchant travels in the output file.

"Algorithmic Analysis"

Direct use of the Tarjan offline LCA algorithm can pass all the data.

"Program Code"

1#include <cstdio>2#include <cstring>3 using namespacestd;4 intg[30010][510],dis[30010],num[30010],point[30010],n,m,ans=0;5 intfa[30010],que[30010][3];6 BOOLbo[30010];7 voidBuildtree (intu) {8      for(intI=1; i<=num[u];i++)9     if(!Bo[g[u][i]]) {Tenbo[g[u][i]]=true; Onedis[g[u][i]]=dis[u]+1; A Buildtree (G[u][i]); -     } - } the intFindintu) { -     returnfa[u]==u?u:fa[u]=find (Fa[u]); - } - voidTarjan (intu) { +bo[u]=true; -      for(intI=1; i<=que[u][0];i++) +     if(Bo[que[u][i]]) ans+=dis[u]+dis[que[u][i]]-2*Dis[find (Que[u][i]); A      for(intI=1; i<=num[u];i++) at     if(!Bo[g[u][i]]) { - Tarjan (G[u][i]); -fa[g[u][i]]=u; -     } - } - intMain () { inMemset (G,0,sizeof(g)); -memset (NUM,0,sizeof(num)); tomemset (DIS,0,sizeof(DIS)); +memset (Point,0,sizeof(point)); -Memset (Bo,0,sizeof(bo)); thememset (que,0,sizeof(que)); *scanf"%d",&n); $      for(intI=1; i<=n;i++) fa[i]=i;Panax Notoginseng      for(intI=1; i<n;i++){ -         intx, y; thescanf"%d%d",&x,&y); +g[x][++num[x]]=y; Ag[y][++num[y]]=x; the     } +bo[1]=true; dis[1]=0; -Buildtree (1); $scanf"%d",&m); $Memset (Bo,0,sizeof(bo)); -      for(intI=1; i<=m;i++) { -scanf"%d",&point[i]); the         if(i==1&&point[i]==1)Continue; -         Else if(i==1&&point[i]!=1){Wuyique[point[i]][++que[point[i]][0]]=1; theque[1][++que[1][0]]=Point[i]; -         } Wu         Else{ -que[point[i]][++que[point[i]][0]]=point[i-1]; Aboutque[point[i-1]][++que[point[i-1]][0]]=Point[i]; $         } -     } -Tarjan (1); -printf"%d", ans); A     return 0; +}

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[Codevs 1036] Business travel