hnu 12029 LCA问题

Ants Colony

Time Limit: 5000ms, Special Time Limit:15000ms, 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 many 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

6 0 8 1 7 1 9 0 3 4 2 4 2 3 5 2 1 4 0 3 2 0 1 2 1 0 0 1 6 0 1000000000 1 1000000000 2 1000000000 3 1000000000 4 1000000000 1 5 0 0

Sample Output

16 20 11 17 1 1 5000000000

Problem Source

ACM ICPC2010 ? Latin American Regional 分析：这题很明显是最小公共祖先问题，比赛时敲模板，敲完一直wa。。。结束才发现悲剧地输出没换行 #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; }