How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15643    Accepted Submission(s): 5942 Problem Description There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.   Input First line is a single integer T(T<=10), indicating the number of test cases.   For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.   Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.   Output For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.   Sample Input 2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1   Sample Output 10 25 100 100  

  题意：有n个房子，n-1条路把它们链接起来，给出每条路的距离，有k个问题，询问两个房间的最小距离。

分析：n个点，n-1条边，组成了一颗树，所以 这是求树上两结点距离的问题，试了一手自己的倍增法求LCA模板。

之前有一篇写过倍增法求LCA 的详解  点击打开链接

AC代码：

#include<stdio.h> #include<string.h> #include<vector> #include<math.h> using namespace std; struct tree { int from,to,w; tree(int ff,int tt,int ww) { from=ff;to=tt;w=ww; } tree(){} }; vector<tree>G; vector<int>V[400200]; int mxdeep; int father[50050][30]; //记录祖先结点编号 int deep[50050]; //记录结点深度 int dis[50050][30]; //记录结点和祖先结点的距离 void addtree(int from,int to,int w) //建边 { G.push_back(tree(from,to,w)); G.push_back(tree(to,from,w)); int e=G.size(); V[from].push_back(e-2); V[to].push_back(e-1); } void init(int n) { memset(father,0,sizeof(father)); memset(deep,0,sizeof(deep)); memset(dis,0,sizeof(dis)); mxdeep=log(n*1.0)/log(2.0); G.clear(); for(int i=0;i<=400000;i++) V[i].clear(); } void dfs(int root) { for(int i=1;i<=mxdeep;i++) { father[root][i]=father[father[root][i-1]][i-1]; dis[root][i]=dis[father[root][i-1]][i-1]+dis[root][i-1]; if(!father[root][i]) //优化，当为0的时候说明已经更新到了根结点位置 break; } for(int i=0;i<V[root].size();i++) { tree &e=G[V[root][i]]; if(e.to!=father[root][0]) //说明e.to是root的儿子 { deep[e.to]=deep[root]+1; father[e.to][0]=root; //建立父子关系 dis[e.to][0]=e.w; //更新到父亲的距离 dfs(e.to); } } } int lca(int u,int v) { int ans=0; if(deep[u]>deep[v]) //让u结点不在v结点下方，方便操作 swap(u,v); for(int i=mxdeep;i>=0;i--) //将 v结点上移，但不能高于u结点 { if(deep[u]<deep[v]&&deep[u]<=deep[father[v][i]]) { // printf("! %d %d %d %d\n",u,v,i,father[v][i]); ans+=dis[v][i]; v=father[v][i]; } } if(v==u) return ans; for(int i=mxdeep;i>=0;i--) { if(father[v][i]!=father[u][i]) //一起移动到共同父亲处 { ans+=dis[v][i]+dis[u][i]; v=father[v][i]; u=father[u][i]; } } if(v!=u) ans+=dis[v][0]+dis[u][0]; // 如果没有移动到同一处，可能是 1->u 1->v ，那么向上移动一个位置就可以了 return ans; } int main() { int T; scanf("%d",&T); while(T--) { int n,k; scanf("%d%d",&n,&k); init(n); int root; for(int i=0;i<n-1;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); father[v][0]=u; dis[v][0]=w; addtree(u,v,w); if(father[u][0]==0) root=u; } deep[root]=1; dfs(root); //建树 for(int i=0;i<k;i++) { int u,v; scanf("%d%d",&u,&v); printf("%d\n",lca(u,v)); } } }