"Business travel" problem Solving report

"Business travel" problem Solving report

By MPs

"Title 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 description"

the first line in the input file has an integer n, 1<=n<=30 000, 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 description"

Export the shortest time a merchant travels in the output file

"Input Sample"

5
1 2
1 5
3 5
4 5
4
1
3
2
5

"Output Example"

7

Analysis

Mathematical modeling: Build a tree, find the shortest path on each tree, algorithm LCA

Shortest path of two points =d (x) +d (y) -2*d (LCA (x, y))

D is the shortest path to the root point

Obviously, that's the depth of the point.

It can be calculated by multiplying, and then it's OK to simulate it directly.

Code

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5 using namespacestd;6 7 Const intmaxn=70001;8 9 intdeep[maxn],p[maxn][ -];Ten intn,q; One  A structlist{ -     intTo,next; - }E[MAXN]; the intHead[maxn],cnt=0; -  - voidAddedge (intUintv) { -cnt++; +e[cnt].to=v; -e[cnt].next=Head[u]; +head[u]=CNT; A } at  - voidinit () { -Freopen ("trip.in","R", stdin); -Freopen ("Trip.out","W", stdout); -scanf"%d",&n); -   inti,u,v; in    for(i=1; i<n;i++){ -scanf"%d%d",&u,&v); to Addedge (u,v); + Addedge (v,u); -   } the } *  $ intLcaintUintv) {Panax Notoginseng     if(deep[u]<Deep[v]) swap (U,V); -     intc=deep[u]-deep[v],i; the      for(i=0; i<= at; i++) +       if((1<<i) &c) Au=P[u][i]; the      for(i= at; i>=0; i--) +       if(p[u][i]!=P[v][i]) { -u=P[u][i]; $v=P[v][i]; $       } -     if(U==V)returnu; -     Else returnp[u][0];  the } - Wuyi voidDfsintu) { the     inti; -      for(i=1; i<= at; i++){ Wu       if(deep[u]< (1<<i)) Break; -p[u][i]=p[p[u][i-1]][i-1]; About     } $      for(i=head[u];i;i=e[i].next) -       if(!Deep[e[i].to]) { -deep[e[i].to]=deep[u]+1; -p[e[i].to][0]=u; A DFS (e[i].to); +       } the } -  $ voidsolve () { thescanf"%d",&q); the     inti,u,v,ans=0, LCA; the      for(i=1; i<=n;i++) the       if(!Deep[i]) { -p[i][0]=i; indeep[i]=1; the DFS (i); the       } Aboutscanf"%d",&u); the      for(i=2; i<=q;i++){ thescanf"%d",&v); theLca=LCA (U,V); +ans+=deep[u]+deep[v]-2*Deep[lca]; -u=v; the     }Bayiprintf"%d", ans); the } the  - intMain () { - init (); the solve (); the     return 0; the}

"Business travel" problem Solving report