[最近公共祖先] POJ 1330 Nearest Common Ancestors
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 27316 | Accepted: 14052 |
Description
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
Output
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
Source
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int num,frist[20010],node[20010],deep[20010],ancestor[20010][17],n;
struct mp
{
int to,next;
}map[20010];
void init()
{
num=0;
memset(ancestor,0,sizeof(ancestor));
memset(frist,0,sizeof(frist));
memset(deep,0,sizeof(deep));
memset(node,0,sizeof(node));
memset(map,0,sizeof(map));
}
void add(int x,int y)
{
++num;
map[num].to=y;
map[num].next=frist[x];frist[x]=num;
}
void build(int v)
{
int i;
for(i=frist[v];i;i=map[i].next)
{
if(!deep[map[i].to])
{
ancestor[map[i].to][0]=v;
deep[map[i].to]=deep[v]+1;
build(map[i].to);
}
}
}
void init_ancestor()
{
int i,j;
for(j=1;j<17;++j)
for(i=1;i<=n;++i)
if(ancestor[i][j-1])
ancestor[i][j]=ancestor[ancestor[i][j-1]][j-1];
return;
}
int lca(int a,int b)
{
int i,dep;
if(deep[a]<deep[b]) swap(a,b);
dep=deep[a]-deep[b];
for(i=0;i<17;++i) if((1<<i)&dep) a=ancestor[a][i];
if(a==b) return a;
for(i=16;i>=0;--i)
{
if(ancestor[a][i]!=ancestor[b][i])
{
a=ancestor[a][i];
b=ancestor[b][i];
}
}
return ancestor[a][0];
}
int main()
{
int T,i,v,w,roof,qv,qw;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d",&n);
for(i=0;i<n-1;++i)
{
scanf("%d%d",&v,&w);
add(v,w);add(w,v);
ancestor[w][0]=v;
if(!ancestor[v][0]) roof=v;//找根,根不同答案是不同的。
}
deep[roof]=1;
build(roof);
init_ancestor();
scanf("%d%d",&qv,&qw);
printf("%d\n",lca(qv,qw));
}
return 0;
}
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