How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15664 Accepted Submission(s): 5950
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
学模板
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int cases,n,q; 7 int a,b,c; 8 struct data{ 9 int next,to,dis; 10 }edge[100010]; 11 int cnt,head[50010],father[50010],w[50010]; 12 bool check[50010]; 13 void add_edge(int start,int end,int d){ 14 edge[++cnt].next=head[start]; 15 edge[cnt].to=end; 16 edge[cnt].dis=d; 17 head[start]=cnt; 18 } 19 struct dt{ 20 int nt,t,num; 21 }eg[40010]; 22 int ct,hd[20010],lca[20010]; 23 void add_ask(int start,int end,int nm){ 24 eg[++ct].nt=hd[start]; 25 eg[ct].t=end; 26 eg[ct].num=nm; 27 hd[start]=ct; 28 } 29 int find(int x){ 30 if(x!=father[x]) father[x]=find(father[x]); 31 return father[x]; 32 } 33 void dfs(int u){ 34 check[u]=1; 35 father[u]=u; 36 for(int i=hd[u];i;i=eg[i].nt) 37 if(check[eg[i].t]) lca[eg[i].num]=find(eg[i].t); 38 for(int i=head[u];i;i=edge[i].next) 39 if(!check[edge[i].to]){ 40 w[edge[i].to]=w[u]+edge[i].dis; 41 dfs(edge[i].to); 42 father[edge[i].to]=u; 43 } 44 } 45 int main(){ 46 scanf("%d",&cases); 47 for(int tt=1;tt<=cases;tt++){ 48 memset(check,0,sizeof(check)); 49 memset(head,0,sizeof(head)); 50 memset(hd,0,sizeof(hd)); 51 memset(edge,0,sizeof(edge)); 52 memset(eg,0,sizeof(eg)); 53 memset(father,0,sizeof(father)); 54 memset(lca,0,sizeof(lca)); 55 memset(w,0,sizeof(w)); 56 ct=0; 57 cnt=0; 58 scanf("%d%d",&n,&q); 59 for(int i=1;i<n;i++){ 60 scanf("%d%d%d",&a,&b,&c); 61 add_edge(a,b,c); 62 add_edge(b,a,c); 63 } 64 for(int i=1;i<=q;i++){ 65 scanf("%d%d",&a,&b); 66 add_ask(a,b,i); 67 add_ask(b,a,i); 68 } 69 dfs(1); 70 for(int i=1;i<=2*q;i+=2){ 71 a=eg[i].t; 72 b=eg[i+1].t; 73 c=eg[i].num; 74 printf("%d ",w[a]+w[b]-2*w[lca[c]]); 75 } 76 } 77 return 0; 78 }