标签：2586 HDU far house deep int each answer line
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18652 Accepted Submission(s):
7268
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
Source ECJTU 2009 Spring Contest
Recommend lcy | We have carefully selected several similar problems for you: 3486 2874 2888 3234 2818 带权的LCA问题 我们用g[i]表示i号节点走到根的权值 那么两个点之间的路径权值为,$g[x]+g[y]-2*g[LCA(x,y)]$ 大概是这个样子 被圆圈出来的是需要减去的
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 node
15 {
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 }
标签：2586,HDU,far,house,deep,int,each,answer,line 来源： https://blog.51cto.com/u_15239936/2868813
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