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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 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 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,..., 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 Source |
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只有一条询问 直接遍历。。
遍历两个节点的所有祖先 发现第一个共同的就是结果
#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;
const int N=10005;
int fa[N];
bool has[N];
int n;
int find(int x,int y)
{
memset(has,false,sizeof(has));
while(fa[x])
{
has[x]=true;
x=fa[x];
}
has[x]=true;
while(fa[y])
{
if(has[y])
break;
y=fa[y];
}
return y;
}
int main()
{
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
memset(fa,0,sizeof(fa));
scanf("%d",&n);
for(int i=0;i<n-1;i++)
{
int a,b;
scanf("%d %d",&a,&b);
fa[b]=a;
}
int x,y;
scanf("%d %d",&x,&y);
printf("%d\n",find(x,y));
}
return 0;
}