Hnu 12029 LCA Problem

Ants colony

Time limit:5000 ms,Special time limit:15000 ms,Memory limit:65536kb

Total submit users:13,Accepted users:6

Problem 12029:No special judgement

Problem description

A group of ants is really proud because they built a magnificent and large colony. however, the enormous size has become a problem, because tables ants do not know the path between some parts of the colony. they desperately need your help! The colony of ants was made as a series of N anthills, connected by tunnels. the ants, obsessive as they are, numbered the anthills sequentially as they built them. the first anthill, numbered 0, did not require any tunnel, but for each of the subsequent anthills, 1 through n-1, the ants also built a single tunnel that connected the new anthill to one of the existing anthills. of course, this tunnel was enough to allow any ant to go to any previously built anthill, possibly passing through other anthills in its path, so they did not bother making extra tunnels and continued building more anthills. Your job is, given the structure of the colony and a set of queries, calculate for each query the shortest path between pairs of anthills. the length of a path is the sum of the lengths of all tunnels that need to be traveled.

Input

Each test case is given using several lines. the first line contains an integer n representing the number of anthills in the colony (2 ≤ n ≤105 ). each of the next n-1 lines contains two integers that describe a tunnel. line I, for 1 ≤ I ≤ n-1, contains AI and Li, indicating that anthill I was connected directly to anthill ai with a tunnel of length Li (0 ≤ AI ≤ I-1 and 1 ≤ Li ≤ 109 ). the next line contains an integer Q representing the number of queries that follow (1 ≤q ≤ 105 ). each of the next Q lines describes a query and contains two distinct integers S and T (0 ≤ S, T ≤ n-1), representing respectively the source and target anthills. The last test case is followed by a line containing one zero.

Output

For each test case output a single line with Q integers, each of them being the length of a shortest path between the pair of Anthills of a query. write the results for each query in the same order that the queries were given in the input.

Sample Input

60 81 71 90 34 242 35 21 40 320 121 00 160 10000000001 10000000002 10000000003 10000000004 100000000015 00

Sample output

16 20 11 171 15000000000

Problem Source

ACM icpc2010? Latin American Regional Analysis: This question is obviously a problem of the least common ancestor. During the competition, the template is knocked out, and the question is always Wa... When it is over, we find that there is no line feed output in a tragic way. #include<cstdio>#include<cstring>using namespace std;const int maxn=111111;const int maxk=111111;struct edge{ int v,next; __int64 w;}g[maxn*2],q[maxk*2];int headg[maxn],headq[maxn],parent[maxn],pos1,pos2,n,m,k,a,b,c;__int64 ans[maxk],dis[maxn];bool vis[maxn];inline void add1(int a,int b,int c){ g[pos1].next=headg[a]; g[pos1].v=b; g[pos1].w=c; headg[a]=pos1++;}inline void add2(int a,int b,int c){ q[pos2].next=headq[a]; q[pos2].v=b; q[pos2].w=c; headq[a]=pos2++;}int find(int a){ if(parent[a]==-1)return a; return parent[a]=find(parent[a]);}void tarjan(int u){ vis[u]=1; int v,i; for(i=headq[u]; i; i=q[i].next) if(vis[v=q[i].v])ans[q[i].w]=dis[u]+dis[v]-2*dis[find(v)]; for(i=headg[u]; i; i=g[i].next) if(!vis[v=g[i].v]) { dis[v]=dis[u]+g[i].w; tarjan(v); parent[v]=u; }}int main(){ while(scanf("%d",&n),n) { memset(headg,0,sizeof(headg)); memset(headq,0,sizeof(headq)); memset(vis,0,sizeof(vis)); for(int i=0; i<=n; ++i)parent[i]=-1; pos1=pos2=1; for(int i=1; i<n; ++i) { scanf("%d%d",&b,&c); add1(i,b,c); add1(b,i,c); } scanf("%d",&m); for(int i=1; i<=m; ++i) { scanf("%d%d",&a,&b); add2(a,b,i); add2(b,a,i); } dis[0]=0; tarjan(0); for(int i=1; i<m; ++i)printf("%I64d ",ans[i]); printf("%I64d\n",ans[m]); } return 0;}