How far away ？
http://www.cnblogs.com/caiyishuai/p/8572859.html
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20491 Accepted Submission(s):
8010
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.
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
Source
Recommend
这一道题目意思是说，村庄之间有路可达，给你N个节点，N-1条路，然后M组查询，查询两个节点之间的距离。
N个节点N-1条边，那么就符合树的定义。所以题目给的就是一个树，就是求树上两个节点的距离。这一道题目可以和LCA联系起来，求两个节点（a和b）的距离。两个节点必然由一个公共点连接起来，这个点就是LCA（最近公共祖先c）
那么求距离就可以转换为a到根节点的距离+b到根节点的距离—c到根节点的距离—c到根节点的距离。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<functional> #define N 100000+10 using namespace std; int n,m; struct node { int to,next,cost; }e[N]; int cnt; int fa[20][N]; int head[N],depth[N],dis[N]; void init() { memset(head,-1,sizeof head); memset(depth,0,sizeof depth); memset(dis,0,sizeof dis); cnt=0; } void addedge(int u,int v,int w)//建图过程，建双向边 { e[cnt].to=v; e[cnt].cost=w; e[cnt].next=head[u]; head[u]=cnt++; } void DFS(int u,int f)//遍历树 { fa[0][u]=f; for(int i=head[u];~i;i=e[i].next)//遍历所有相连的边 { int To=e[i].to; if(To!=f)//去掉以后MLE，可能是递归求的过程中太多临时变量 {//建树过程建双向边，会出现to=f的情况，去掉以后会陷入无限递归中 dis[To]=dis[u]+e[i].cost;//更新距离 depth[To]=depth[u]+1;//更新深度 DFS(To,u); } } } void solve() { depth[1]=1;//题目给的是一个树 dis[1]=0;//无论怎么样的树，都可以把1视为根节点 DFS(1,-1); for(int i=1;i<20;i++)//树上倍增 for(int j=1;j<=n;j++) fa[i][j]=fa[i-1][fa[i-1][j] ]; } int LCA(int u,int v)//求最近公共祖先 { if(depth[u]>depth[v])//保证V的深度比较大 swap(u,v); for(int i=0;i<20;i++)//倍增到深度相同 if((depth[v]-depth[u])>>i&1)//二进制特性，一定能跳到深度相同 v=fa[i][v]; if(u==v) return u; for(int i=19;i>=0;i--)//两者同时倍增 { if(fa[i][u]!=fa[i][v]) { u=fa[i][u]; v=fa[i][v]; } } return fa[0][v]; } int main() { int i,t; int a,b,c; while(scanf("%d",&t)!=EOF) { while(t--) { init(); scanf("%d%d",&n,&m); for(i=1;i<n;i++) { scanf("%d%d%d",&a,&b,&c); addedge(a,b,c); addedge(b,a,c); } solve(); for(i=1;i<=m;i++) { scanf("%d%d",&a,&b); int ans=dis[a]+dis[b]-2*dis[LCA(a,b)]; printf("%d ",ans); } } return 0; } }