标签：ase following iostream tin ati like can for sam
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.InputFirst 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.OutputFor 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倍增
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 5 using namespace std; 6 7 const int maxn=40005; 8 const int LOG = 20; 9 int par[maxn][LOG],dep[maxn]; 10 int head[maxn],dis[maxn]; 11 int cnt=0; 12 13 struct EDGE 14 { 15 int v,w,next; 16 }edge[maxn*2]; 17 18 void addedge(int u,int v,int w) 19 { 20 edge[cnt].v=v; 21 edge[cnt].w=w; 22 edge[cnt].next=head[u]; 23 head[u]=cnt++; 24 } 25 26 void dfs(int u,int fa,int depth){ 27 dep[u]=depth; 28 if(u==1) 29 { 30 for(int i=0;i<LOG;i++) par[u][i]=u; 31 } 32 else 33 { 34 par[u][0]=fa; 35 for(int i=1;i<LOG;i++) par[u][i]=par[par[u][i-1]][i-1]; 36 } 37 for(int i=head[u];i!=-1;i=edge[i].next) 38 { 39 int q=edge[i].v; 40 if(q==fa) continue; 41 dis[q]=dis[u]+edge[i].w; 42 dfs(q,u,depth+1); 43 } 44 } 45 46 int up(int x,int step){ 47 for(int i=0;i<LOG;i++) 48 if(step&(1<<i)) 49 x=par[x][i]; 50 return x; 51 } 52 53 int lca(int u,int v){ 54 if(dep[u]<dep[v]) 55 swap(u,v); 56 u=up(u,dep[u]-dep[v]); 57 if(u==v) return u; 58 for(int i=LOG-1;i>=0;i--) 59 { 60 if(par[u][i]!=par[v][i]) 61 { 62 u=par[u][i]; 63 v=par[v][i]; 64 } 65 } 66 return par[u][0]; 67 } 68 69 int main() 70 { 71 int T,n,m; 72 cin>>T; 73 while(T--) 74 { 75 cin>>n>>m; 76 cnt=0; 77 memset(head,-1,sizeof(head)); 78 int x,y,z; 79 for(int i=0;i<n-1;i++) 80 { 81 scanf("%d%d%d",&x,&y,&z); 82 addedge(x,y,z); 83 addedge(y,x,z); 84 } 85 dis[1]=0; 86 dfs(1,-1,0); 87 for(int i=0;i<m;i++) 88 { 89 scanf("%d%d",&x,&y); 90 int ans=dis[x]+dis[y]-2*dis[lca(x,y)]; 91 printf("%d\n",ans); 92 } 93 } 94 95 return 0; 96 }
标签：ase following iostream tin ati like can for sam
原文地址：http://www.cnblogs.com/xibeiw/p/7421257.html
