POJ 1330 Nearest Common Ancestors
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 14698 | Accepted: 7839 |
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
题目大意:求树节点的最近公共祖先。
解题方法:这种题解题方法很多,我在这里用的是回溯,直接从要查找的节点不断的找父节点。
#include <stdio.h>
#include <iostream>
#include <vector>
#include <string.h>
using namespace std; typedef struct
{
int parent;
bool bvisted;
}UFSTree; UFSTree Tree[]; void MakeSet(int n)
{
for (int i = ; i <= n; i++)
{
Tree[i].parent = i;
Tree[i].bvisted = false;
}
} void LCA(int x, int y)
{
Tree[x].bvisted = true;
x = Tree[x].parent;
while(Tree[x].parent != x)
{
Tree[x].bvisted = true;
x = Tree[x].parent;
}
while(Tree[y].parent != y)
{
if (Tree[y].bvisted == true)
{
break;
}
y = Tree[y].parent;
}
printf("%d\n", y);
} int main()
{
int n, nCcase, father, son, x, y;
scanf("%d", &nCcase);
while(nCcase--)
{
scanf("%d", &n);
MakeSet(n);
for (int i = ; i < n; i++)
{
scanf("%d%d", &father, &son);
Tree[son].parent = father;
}
scanf("%d%d", &x, &y);
LCA(x, y);
}
}
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