HDU 2586 How far away ?, Hdu2586

HDU 2586 How far away ?, Hdu2586

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 18652 Accepted Submission (s): 7268

Problem DescriptionThere 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 Input23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output1025100100

 

SourceECJTU 2009 Spring Contest

 

Recommendlcy | We have carefully selected several similar problems for you: 3486 2874 2888 3234 2818 weighted LCA problem we use g [I] to indicate the weight of the I node to the root. Then the path weight between the two points is, $ g [x] + g [y]-2 * g [LCA (x, y)] $ this is probably what needs to be subtracted from the circle.

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int MAXN=1e5+10; 6 inline int read() 7 { 8     char c=getchar();int x=0,f=1; 9     while(c<'0'||c>'9')    {if(c=='-')    f=-1;c=getchar();}10     while(c>='0'&&c<='9')    x=x*10+c-48,c=getchar();return x*f;11 }12 int n,m,S=1; 13 int f[MAXN][21],deep[MAXN],g[MAXN];14 struct node15 {16     int u,v,w,nxt;17 }edge[MAXN];18 int head[MAXN];19 int num=1;20 inline void add_edge(int x,int y,int z)21 {22     edge[num].u=x;23     edge[num].v=y;24     edge[num].w=z;25     edge[num].nxt=head[x];26     head[x]=num++;27 }28 void dfs(int now)29 {30     for(int i=head[now];i!=-1;i=edge[i].nxt)31         if(deep[edge[i].v]==0)32         {33             deep[edge[i].v]=deep[now]+1;34             f[edge[i].v][0]=now;35             g[edge[i].v]=g[now]+edge[i].w;36             dfs(edge[i].v);37         }38             39 }40 inline void pre()41 {42     for(int i=1;i<=19;i++)43         for(int j=1;j<=n;j++)44             f[j][i]=f[f[j][i-1]][i-1];45 }46 inline int LCA(int x,int y)47 {48     if(deep[x]<deep[y])    swap(x,y);49     for(int i=19;i>=0;i--)50         if(deep[f[x][i]]>=deep[y])51             x=f[x][i];52     if(x==y)    return x;53 54     for(int i=19;i>=0;i--)55         if(f[x][i]!=f[y][i])56             x=f[x][i],y=f[y][i];57     return f[x][0];58 }59 int main()60 {61     int T=read();62     while(T--)63     {64         n=read();m=read();65         memset(head,-1,sizeof(head));num=1;66         memset(f,0,sizeof(f));67         memset(deep,0,sizeof(deep));68         for(int i=1;i<=n-1;i++)69         {70             int x=read(),y=read(),z=read();71             add_edge(x,y,z);72             add_edge(y,x,z);73         }74         deep[S]=1;75         dfs(S);pre();76         while(m--)77         {78             int x=read(),y=read();79             printf("%d\n",g[x]+g[y]-2*g[LCA(x,y)]);80         }    81     }82     83     return 0;84 }