POJ 1308

Is It A Tree?

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43024		Accepted: 13980

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. 

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.

题目：给定有向边，判断这些边是否构成树。

并查集模板，在判断是否有两条边指向同一个节点时，只需判断两个节点是否在一个集合中，因为某节点若已被指向一次，就加入到集合中，第二次肯定属于同一集合。

程序：

#include
#include
#include
#include
#include
#include
#define INF 2e8
#define MAXN 50010
using namespace std;

int parent[MAXN], ranks[MAXN], cnt[MAXN], noden[MAXN], nodem[MAXN], vis[MAXN];

int findRoot(int v) {
    while(v != parent[v]) {
        parent[v] = parent[parent[v]];
        v = parent[v];
    }
    return v;
}

void unions(int a, int b) {
    int aRoot = findRoot(a);
    int bRoot = findRoot(b);
    if (aRoot == bRoot) return;
    if (ranks[aRoot] < ranks[bRoot]) parent[aRoot] = bRoot;
    else if (ranks[aRoot] > ranks[bRoot]) parent[bRoot] = aRoot;
    else {
        parent[aRoot] = bRoot;
        ranks[bRoot]++;
    }
}

void init() {
    for (int i = 1;i <= MAXN;i++){
         parent[i] = i;
         ranks[i] = 1;
    }
}

int same(int a, int b) {
    return findRoot(a) == findRoot(b);
}

int main()
{
    int n, m, a, b, sum = 1, x = 0;
    bool over = false;
    int cas = 1;
    while(1) {
        bool OK = true;
        set s;
        init();
        while(1) {
            scanf("%d%d", &a, &b);
            if (a == b && b == 0) break;
            if (a == b && a == -1) {
                over = true;
                break;
            }
            s.insert(a);
            s.insert(b);
            if (!same(a,b) && OK) unions(a, b);
            else {
                OK = false;
                continue;
            }
        }
        if (over) break;
        set::iterator it;
        it = s.begin();
        int root = findRoot(*it);
        for (;it != s.end();it++) {
            if (root != findRoot(*it)) {
                OK = false;
                break;
            }
        }
        if (OK) printf("Case %d is a tree.\n", cas);
        else printf("Case %d is not a tree.\n", cas);
        cas++;
    }
    return 0;
}