poj 1308 Is It A Tree?
2012-05-13 10:16
302 查看
Is It A Tree?
Description
A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.
![](http://poj.org/images/1308_1.jpg)
In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.
Input
The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.
Output
For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).
Sample Input
Sample Output
并查集的应用,注意空树也是树。。。。。
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 14525 | Accepted: 5007 |
A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.
![](http://poj.org/images/1308_1.jpg)
In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.
Input
The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.
Output
For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).
Sample Input
6 8 5 3 5 2 6 4 5 6 0 0 8 1 7 3 6 2 8 9 7 5 7 4 7 8 7 6 0 0 3 8 6 8 6 4 5 3 5 6 5 2 0 0 -1 -1
Sample Output
Case 1 is a tree. Case 2 is a tree. Case 3 is not a tree.
#include<iostream> using namespace std; struct node { int x, y; }edge[1000]; int father[1000]; int num=0; void make_set() { for(int i=0;i<1000;i++) father[i] = i; } int find_father(int x) { if(x != father[x]) return find_father(father[x]); return x; } void union_set(int a, int b) { father[b] = a; } int main() { int a, b; int flag=0; int i; int cases = 1; freopen("e:\\data.txt", "r", stdin); freopen("e:\\out.txt", "w", stdout); make_set(); while(cin>>a>>b) { if(a==-1&&b==-1) break; if(a==0&&b==0) { if(num==0) cout<<"Case "<<cases<<" is a tree."<<endl; else if(flag==0) { int f = find_father(edge[0].x); for(i=1;i<num;i++) { if(find_father(edge[i].x)!=f) break; } if(i==num) cout<<"Case "<<cases<<" is a tree."<<endl; else cout<<"Case "<<cases<<" is not a tree."<<endl; } else { cout<<"Case "<<cases<<" is not a tree."<<endl; } num=0; flag=0; make_set(); cases++; continue; } edge[num].x=a; edge[num].y=b; num++; int p=find_father(a); int q=find_father(b); if(p==q || q!=b) { flag=1; } else { union_set(a, b); } } return 0; }
并查集的应用,注意空树也是树。。。。。
相关文章推荐
- POJ-1308-Is It A Tree?
- POJ 1308 Is It A Tree?
- POJ1308 Is It A Tree?(树,并查集)
- 【并查集】hdu 1325 Is It A Tree? 或 poj 1308 Is It A Tree?
- poj 1308 is it a tree?(图的性质||基础并查集) (同hdu 1272 小希的迷宫 )
- [POJ 1308]Is It A Tree?(并查集判断图是否为一棵有根树)
- POJ 1308 Is it a tree(并查集)
- POJ-1308 Is It A Tree?
- poj 1308 Is It A Tree?
- poj 1308 Is It A Tree?(并查集)
- POJ 1308 Is It A Tree?(并查集)
- POJ 1308 Is It A Tree? && NYOJ 129 (树的判定+并查集)
- poj 1308 Is It A Tree?(并查集)
- poj 1308 - Is It A Tree?(并查集)
- poj 1308 Is It A Tree?
- HDU 1325&&POJ 1308 Is It A Tree? 并查集判断能否树(能否形成环)
- POJ 1308 Is It A Tree?
- HDU 1325 POJ 1308 Is It ATree?(并查集)
- POJ 1308 Is It A Tree?
- POJ 1308 Is It A Tree?(并查集)