Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to change this situation.
In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.
Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.
Input
The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li < 100) indicating there's a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the indices of the three regions to be checked.
Process to the end of file.
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Output
Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.
Output a blank line between each test cases.
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Sample Input
4 0 1 1 0 2 1 0 3 1 2 1 2 3 0 1 2 5 0 1 1 0 2 1 1 3 1 1 4 1 2 0 1 2 1 0 3
<b< dd="">
Sample Output
3 2 2 2
题解：Orz,PE了好几发，格式真的。。。有毒
题目让求3点间的最短距离；我们用LCA求任意两点间的最短距离，然后将三个结果求和然后除以二即可；画图很容易理解；
参考代码为：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<cstdlib> 6 #include<algorithm> 7 #include<queue> 8 #include<stack> 9 #include<set> 10 #include<vector> 11 #include<map> 12 using namespace std; 13 typedef long long LL; 14 const int N=1e5+10,maxn=500+100,inf=0x3f3f3f3f; 15 int a,b,c; 16 struct edge{ 17 int to,Next,w; 18 }e[N]; 19 int head[N],dis[N]; 20 int cnt,depth[N],fa[20][N]; 21 void add(int u,int v,int w) 22 { 23 e[cnt].to=v; 24 e[cnt].w=w; 25 e[cnt].Next=head[u]; 26 head[u]=cnt++; 27 } 28 void dfs(int u,int f) 29 { 30 fa[0][u]=f; 31 for(int i=head[u];~i;i=e[i].Next) 32 { 33 int To=e[i].to; 34 if(To!=f) 35 { 36 depth[To]=depth[u]+1; 37 dis[To]=dis[u]+e[i].w; 38 dfs(To,u); 39 } 40 } 41 } 42 void work(int n) 43 { 44 depth[1]=1;dis[1]=0; 45 dfs(1,-1); 46 for(int i=1;i<20;i++) 47 for(int j=1;j<=n;j++) 48 fa[i][j]=fa[i-1][fa[i-1][j]]; 49 } 50 int lca(int x,int y) 51 { 52 if(depth[x]>depth[y]) swap(x,y); 53 for(int i=0;i<20;i++) 54 if((depth[y]-depth[x])>>i&1) y=fa[i][y]; 55 if(x==y) return x; 56 for(int i=19;i>=0;i--) 57 { 58 if(fa[i][x]!=fa[i][y]) 59 { 60 x=fa[i][x]; 61 y=fa[i][y]; 62 } 63 } 64 return fa[0][x]; 65 } 66 int getdis(int x,int y) 67 { 68 return dis[x]+dis[y]-2*dis[lca(x,y)]; 69 } 70 int main() 71 { 72 int n,temp=0; 73 while(~scanf("%d",&n)) 74 { 75 if(temp) puts(""); 76 temp=1; 77 cnt=0; 78 memset(head,-1,sizeof head); 79 for(int i=1; i<n; i++) 80 { 81 scanf("%d%d%d",&a,&b,&c); 82 a++,b++; 83 add(a,b,c); 84 add(b,a,c); 85 } 86 work(n); 87 int k; 88 scanf("%d",&k); 89 while(k--) 90 { 91 scanf("%d%d%d",&a,&b,&c); 92 a++,b++;c++; 93 int ans=getdis(b,a)+getdis(c,b)+getdis(a,c); 94 printf("%d ",ans/2); 95 } 96 } 97 return 0; 98 }