NYOJ129 树的判定 【并查集】

标签：nyoj129

描述

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. 

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.

输入

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.

The number of test cases will not more than 20,and the number of the node will not exceed 10000.
The inputs will be ended by a pair of -1.

输出

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).

样例输入

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

样例输出

Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

来源

POJ

上传者

张云聪

题意：判断一个有向图是否是树。

题解：如果一个图是树，那么必须满足以下情况： 

1、树的数量不能大于1，空树也是树；

2、节点入度数不能大于1；

3、不能成环，比如一棵树的叶子节点指向根节点就是非法的；

4、自环是非法的。

#include <stdio.h>
#include <string.h>

#define maxn 10010

int pre[maxn];
bool vis[maxn];

int unionFind(int k){
	int a = k, b;
	while(pre[k] != -1) k = pre[k];
	while(a != k){
		b = pre[a];
		pre[a] = k;
		a = b;
	}
	return k;
}

int main() {
	// freopen("stdin.txt", "r", stdin);
	memset(pre, -1, sizeof(pre));
	int u, v, cas = 1, ok = 1, count = 0;
	while(scanf("%d%d", &u, &v) != EOF) {
		if(u < 0) break;
		if(!(u | v)) {
			printf("Case %d ", cas++);
			if(count > 1) ok = 0;
			if(ok) printf("is a tree.\n");
			else printf("is not a tree.\n");
			memset(pre, -1, sizeof(pre));
			memset(vis, 0, sizeof(vis));
			count = 0; ok = 1; continue;
		}
		if(!ok) continue;

		if(!vis[u]) {
			vis[u] = 1; ++count;
		}
		if(!vis[v]) {
			vis[v] = 1; ++count;
		}
		if(pre[v] != -1 || u == v) {
			ok = 0; continue;
		}
		u = unionFind(u);
		if(u == v) {
			ok = 0; continue;
		}
		pre[v] = u; --count;
	}
	return 0;
}

NYOJ129 树的判定 【并查集】

标签：nyoj129

原文地址：http://blog.csdn.net/chang_mu/article/details/40161035