HDU 2586 How Far Away? LCA水题

 

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

 

 

 

 

题意：给出一棵树，m个询问

每次询问给出a,b

问a,b的距离。

 

siz[i]  表示i到根节点的距离

 

思路：先求出lca=LCA(a,b)

 

则距离 = siz[a]+siz[b]-2*siz[lca]

 

 

 

  1 #include<cstdio>

  2 #include<cstring>

  3 #include<algorithm>

  4 

  5 using namespace std;

  6 

  7 const int maxn=40000+5;

  8 

  9 struct Edge

 10 {

 11     int to,next,w;

 12 }edge[maxn<<1];

 13 int head[maxn];

 14 int tot;

 15 int p[maxn][24];

 16 int siz[maxn];

 17 int dep[maxn];

 18 

 19 void init(int n)

 20 {

 21     tot=1;

 22     memset(head,-1,sizeof(head));

 23     memset(siz,0,sizeof(siz));

 24     memset(dep,0,sizeof(dep));

 25 

 26     for(int i=1;i<=n;i++)

 27         for(int j=0;j<24;j++)

 28             p[i][j]=-1;

 29 }

 30 

 31 void addedge(int u,int v,int w)

 32 {

 33     edge[tot].to=v;

 34     edge[tot].w=w;

 35     edge[tot].next=head[u];

 36     head[u]=tot++;

 37 }

 38 

 39 void dfs(int n,int u,int fa)

 40 {

 41     for(int i=head[u];~i;i=edge[i].next)

 42     {

 43         int v=edge[i].to;

 44         int w=edge[i].w;

 45         if(!dep[v]&&v!=fa)

 46         {

 47             dep[v]=dep[u]+1;

 48             p[v][0]=u;

 49             siz[v]=siz[u]+w;

 50             dfs(n,v,u);

 51         }

 52     }

 53 }

 54 

 55 void init_lca(int n)

 56 {

 57     for(int j=1;(1<<j)<=n;j++)

 58     {

 59         for(int i=1;i<=n;i++)

 60         {

 61             if(p[i][j-1]!=-1)

 62             {

 63                 p[i][j]=p[p[i][j-1]][j-1];

 64             }

 65         }

 66     }

 67 }

 68 

 69 int solve(int n,int u,int v)

 70 {

 71     if(u==v)

 72         return 0;

 73     if(dep[u]<dep[v])

 74     {

 75         swap(u,v);

 76     }

 77 

 78     int init_u=u;

 79     int init_v=v;

 80 

 81     int cnt;

 82     for(cnt=0;(1<<cnt)<=dep[u];cnt++)

 83         ;

 84     cnt--;

 85 

 86     for(int j=cnt;j>=0;j--)

 87     {

 88         if(dep[u]-(1<<j)>=dep[v])

 89             u=p[u][j];

 90     }

 91 

 92     if(u==v)

 93         return (siz[init_u]-siz[u]);

 94 

 95     for(int j=cnt;j>=0;j--)

 96     {

 97         if(p[u][j]!=-1&&p[u][j]!=p[v][j])

 98         {

 99             u=p[u][j];

100             v=p[v][j];

101         }

102     }

103     int lca=p[u][0];

104 

105     return (siz[init_u]+siz[init_v]-2*siz[lca]);

106 

107 }

108 

109 int main()

110 {

111     int test;

112     scanf("%d",&test);

113 

114     while(test--)

115     {

116         int n,m;

117         scanf("%d%d",&n,&m);

118 

119         init(n);

120 

121         for(int i=1;i<n;i++)

122         {

123             int u,v,w;

124             scanf("%d%d%d",&u,&v,&w);

125             addedge(u,v,w);

126             addedge(v,u,w);

127         }

128 

129         dfs(n,1,-1);

130         init_lca(n);

131 

132         /*

133         for(int i=1;i<=n;i++)

134             printf("%d\n",siz[i]);

135         */

136 

137         for(int i=1;i<=m;i++)

138         {

139             int u,v;

140             scanf("%d%d",&u,&v);

141             //printf("eee\n");

142             printf("%d\n",solve(n,u,v));

143         }

144     }

145 

146     return 0;

147 }

31ms