HDU 2586 How far away？（lca模板）

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14028    Accepted Submission(s): 5295


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
 
lca模板
 1 #include <iostream>
 2 #include <cstring>
 3 #include <vector>
 4 #define N 40010
 5 using namespace std;
 6 struct node{
 7     int to,len;
 8 };
 9 int n,parent[N],len[N],ans[N];
10 vector<node> edge[N],qedge[N];
11 bool visit[N];
12 void init(){  //初始化
13     for(int i=1;i<=n;i++){
14         edge[i].clear();
15         qedge[i].clear();
16         parent[i]=i;
17         visit[i]=false;
18         len[i]=0;
19     }
20 }
21 int find(int x){
22     if(x==parent[x]){
23         return x;
24     }
25     return parent[x]=find(parent[x]);
26 }
27 void add(int a,int b){
28     a=find(a);
29     b=find(b);
30     if(a!=b){
31         parent[b]=a;
32     }
33 }
34 void lca(int cnt,int sum){
35     visit[cnt]=true;
36     len[cnt]=sum;
37     int size=edge[cnt].size();
38     for(int i=0;i<size;i++){
39         node aa;
40         aa=edge[cnt][i];
41         if(!visit[aa.to]){
42             lca(aa.to,aa.len+sum);
43             add(cnt,aa.to);
44         }
45     }
46     size=qedge[cnt].size();
47     for(int i=0;i<size;i++){
48         node aa;
49         aa=qedge[cnt][i];
50         if(visit[aa.to]){
51             ans[aa.len]=len[cnt]+len[aa.to]-len[find(aa.to)]*2;
52             //cout<<aa.len<<" "<<ans[aa.len]<<endl;
53         }
54     }
55 }
56 int main(){
57     cin.sync_with_stdio(false);
58     int T,m,a,b,c;
59     cin>>T;
60     while(T--){
61         cin>>n>>m;
62         init();
63         for(int i=1;i<n;i++){
64             cin>>a>>b>>c;
65             node aa;
66             aa.to=b;
67             aa.len=c;
68             edge[a].push_back(aa);
69             aa.to=a;
70             edge[b].push_back(aa);
71         }
72         for(int i=1;i<=m;i++){
73             cin>>a>>b;
74             node aa;
75             aa.to=b;
76             aa.len=i;
77             qedge[a].push_back(aa);
78             aa.to=a;
79             qedge[b].push_back(aa);
80         }
81         lca(1,0);
82         for(int i=1;i<=m;i++){
83             cout<<ans[i]<<endl;
84         }
85     }
86     return 0;
87 }

 

posted @ 2016-12-11 13:47  ガ落涙『不變』  阅读(65)  评论(0)  编辑  收藏  举报