Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 793    Accepted Submission(s): 257

 

Problem 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 45 6 0 08 1 7 3 6 2 8 9 7 57 4 7 8 7 6 0 03 8 6 8 6 45 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.

判断一个图是不是树的关键在于1：不存在回路 2： e + 1 = v。注意输入的顶点序列与并查集的输入序列相反
#include<cstdio>using namespace std;const int MAXN = 1e4;int p[MAXN], node[MAXN], ecnt, ncnt;  //node数组表示是否访问过这个顶点，用于统计顶点数目 void init(){for(int i = 0; i < MAXN; i++){p[i] = i;node[i] = 0;}}int find(int x){if(x != p[x]){p[x] = find(p[x]);}return p[x];}bool join(int a, int b){int x = find(a);int y = find(b);if(a == x && y != a){  //a->b成立的条件是a不存在父节点，b到a不连通 if(!node[a]) {ncnt++;node[a] = 1;}if(!node[b]){ncnt++;node[b] = 1;}ecnt++;p[a] = y;return true;}return false;}int main(){int kase = 0, a, b, flag = 1;init();ecnt = ncnt = 0;while(scanf("%d%d", &a, &b) && (a >= 0 || b >= 0)){while(a != 0 && b != 0){if(!join(b, a)) flag = 0;scanf("%d%d", &a, &b);}if(ecnt + 1 != ncnt && ncnt > 0) flag = 0;if(flag) printf("Case %d is a tree.\n", ++kase);else printf("Case %d is not a tree.\n", ++kase);init();ecnt = ncnt = 0;flag = 1;}return 1;} 

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