POJ1330 Nearest Common Ancestors
Description
A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:
In the figure, each node is labeled with
an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an
ancestor of node y if node x is in the path between the root and node y. For
example, node 4 is an ancestor of node 16. Node 10 is also an ancestor of node
16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16.
Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the
ancestors of node 7. A node x is called a common ancestor of two different
nodes y and z if node x is an ancestor of node y and an ancestor of node z.
Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is
called the nearest common ancestor of nodes y and z if x is a common ancestor
of y and z and nearest to y and z among their common ancestors. Hence, the nearest
common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7
than node 8 is.
For other examples, the nearest common ancestor of nodes 2 and 3 is node 10,
the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common
ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor
of z, then the nearest common ancestor of y and z is y.
Write a program that finds the nearest common ancestor of two distinct nodes in
a tree.
Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,..., N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers whose nearest common ancestor is to be computed.
Output
Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.
Sample Input
2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5
Sample Output
4
3
【题意】：T组数据求询问单次LCA裸题，练手用。
1 #include <stdio.h> 2 #include <iostream> 3 #include <cmath> 4 #include <algorithm> 5 #include <string> 6 #include <string.h> 7 using namespace std; 8 int quex,quey; 9 int du[100000+1]; 10 bool v[100000+1]={}; 11 int fa[100000+1]; 12 int ans; 13 struct node 14 { 15 int next;int to; 16 }e[100000+1]; 17 int cnt; 18 int g[100000+1]; 19 int n; 20 bool flag; 21 inline void add(int x,int y) 22 { 23 e[++cnt].next=g[x]; 24 g[x]=cnt; 25 e[cnt].to=y; 26 } 27 28 int getfa(int x) 29 { 30 return (fa[x]==x?x:getfa(fa[x])); 31 } 32 inline void merge(int x,int y) 33 { 34 fa[getfa(x)]=fa[getfa(y)]; 35 } 36 inline void tarjan(int now) 37 { 38 int i; 39 v[now]=1; 40 if (v[quex] && !flag && now==quey) 41 {ans=getfa(quex);flag=1;} 42 if (v[quey] && !flag && now==quex) 43 {ans=getfa(quey);flag=1;} 44 for (i=g[now];i;i=e[i].next) 45 if (!v[e[i].to]) 46 { 47 48 tarjan(e[i].to); 49 fa[e[i].to]=now; 50 } 51 } 52 inline void release() 53 { 54 for (register int i=1;i<=n;i++) fa[i]=i; 55 cnt=0; 56 flag=0; 57 memset(g,0,sizeof (g)); 58 memset(du,0,sizeof(du)); 59 memset(v,0,sizeof(v)); 60 } 61 int main() 62 { 63 int t; 64 scanf("%d",&t); 65 while(t--) 66 { 67 scanf("%d",&n); 68 release(); 69 int root; 70 for (register int i=1;i<=n-1;i++) 71 { 72 int x ,y; 73 scanf("%d%d",&x,&y); 74 du[y]++; 75 add(x,y); 76 } 77 for (register int i=1;i<=n;i++) 78 if (!du[i]) {root=i;break;} 79 scanf("%d%d",&quex,&quey); 80 tarjan(root); 81 cout<<ans<<endl; 82 } 83 return 0; 84 }
1 #include <iostream> 2 #include <string> 3 #include <string.h> 4 #include <cmath> 5 #include <algorithm> 6 #include <stdio.h> 7 using namespace std; 8 int n; 9 int du[100000+1]={}; 10 int deep[100000+1]; 11 int fa[100000+1][21]; 12 int g[100000+1]; 13 int cnt=0; 14 int maxdeep; 15 struct node 16 { 17 int next;int to; 18 }e[100000+1]; 19 int read() 20 { 21 int num=0;int fh=1;char ch=getchar(); 22 while(ch<'0' || ch>'9'){if (ch=='-') fh=-1;ch=getchar();} 23 while(ch<='9' && ch>='0'){num=(num<<1)+(num<<3)+ch-'0';ch=getchar();} 24 return (num*fh); 25 } 26 inline void add(int x,int y) 27 { 28 e[++cnt].next=g[x]; 29 g[x]=cnt; 30 e[cnt].to=y; 31 } 32 void dfs(int u) 33 { 34 for (register int i=g[u];i;i=e[i].next) 35 deep[e[i].to]=deep[u]+1,fa[e[i].to][0]=u,maxdeep=max(maxdeep,deep[e[i].to]),dfs(e[i].to); 36 } 37 void bz() 38 { 39 int i,j; 40 for (j=1;(1<<j)<=maxdeep;j++) 41 for (i=1;i<=n;i++) 42 fa[i][j]=fa[fa[i][j-1]][j-1]; 43 } 44 int query(int u,int v) 45 { 46 int j; 47 if (deep[v]>deep[u]) swap(u,v); 48 for (j=20;j>=0;j--) 49 if (deep[u]-deep[v]>=(1<<j)) u=fa[u][j]; 50 if (u==v) return u; 51 for (j=20;j>=0;j--) 52 if (fa[u][j]!=fa[v][j]) 53 u=fa[u][j],v=fa[v][j]; 54 return fa[u][0]; 55 } 56 inline void release() 57 { 58 memset(g,0,sizeof (g)); 59 cnt=0; 60 memset(du,0,sizeof (du)); 61 memset(deep,0,sizeof (deep)); 62 memset(fa,0,sizeof(fa)); 63 } 64 int main() 65 { 66 int t,i; 67 68 t=read(); 69 while(t--) 70 { 71 n=read(); 72 release(); 73 for (i=1;i<=n-1;i++) 74 { 75 int x=read(),y=read(); 76 add(x,y); 77 du[y]++; 78 } 79 for (i=1;i<=n;i++) 80 if (!du[i]) break; 81 deep[i]=1; 82 dfs(i); 83 bz(); 84 int quex=read(),quey=read(); 85 cout<<query(quex,quey)<<endl; 86 } 87 return 0; 88 }