How far away ?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3653 Accepted Submission(s): 1379
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
题目分析:题目中说有n个房子,有n-1条道路,并且都是联通的,典型的树状图,要求任意两点的距离,可以用线段树做,此处不提,用最短路
n的范围是4000,用一般最短路的方法一定会超时,因此利用树状结构的特点,最好用LCA;
程序:
#include"stdio.h" #include"string.h" #include"iostream" #include"queue" #include"map" using namespace std; #define M 40005 int dis[M],pre[M],head[M],t,sum,use[M],rank[M]; struct st { int u,v,w,next; }edge[M*3]; void init() { t=0; memset(head,-1,sizeof(head)); } void add(int u,int v,int w) { edge[t].u=u; edge[t].v=v; edge[t].w=w; edge[t].next=head[u]; head[u]=t++; } void bfs(int s) { queue<int>q; memset(use,0,sizeof(use)); memset(rank,0,sizeof(rank)); use[s]=1; q.push(s); while(!q.empty()) { int u=q.front(); q.pop(); for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(!use[v]) { use[v]=1; pre[v]=u;//记录父节点; rank[v]=rank[u]+1;//记录层数 dis[v]=edge[i].w;//记录子节点到父节点的距离 q.push(v); } } } } void targan(int a,int b) { if(a==b) return; else if(rank[a]>rank[b]) { sum+=dis[a]; targan(pre[a],b); } else { sum+=dis[b]; targan(a,pre[b]); } }//用深搜的方法求最近公共祖先;sum记录路径长度 int main() { int w,a,b,c,i,m,n; scanf("%d",&w); while(w--) { init(); scanf("%d%d",&n,&m); for(i=1;i<n;i++) { scanf("%d%d%d",&a,&b,&c); add(a,b,c); add(b,a,c); } bfs(1); while(m--) { scanf("%d%d",&a,&b); sum=0; targan(a,b); printf("%d ",sum); } } }