POJ-1308-Is It A Tree?（并查集 判断树）

F - Is It A Tree?
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

POJ 1308
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.

给出各个边的关系判断这些边能否构成树。
充分利用了并查集的思想，不停的给各个边找爹。

我自己写的比较乱，贴下来自网路的代码；
下面的代码来自http://blog.csdn.net/joyforce/article/details/8038755

#include<stdio.h>#include<iostream>#include<algorithm>#include<string>#include<string.h>#include<math.h>using namespace std;//并查集//判断树const int MAX_SIZE = 105;int parent[MAX_SIZE];bool flag[MAX_SIZE];void make_set(){  //初始化    for(int x = 1; x < MAX_SIZE; x ++){        parent[x] = x;        flag[x] = false;    }}int find_set(int x){   //寻找根节点，带路径压缩    if(x != parent[x])        parent[x] = find_set(parent[x]);    return parent[x];}void union_set(int x, int y){  //合并    x = find_set(x);    y = find_set(y);    if(x == y) return;    parent[y] = x;}bool single_root(int n){   //判断是不是只有一个根，条件(1)    int i = 1;    while (i <= n && !flag[i]){        ++i;    }    int root = find_set(i);    while (i <= n){        if (flag[i] && find_set(i) != root){            return false;        }        ++i;    }    return true;}int main(){    int x, y;    bool is_tree = true;    int range = 0;    int idx = 1;    make_set();    while (scanf("%d %d", &x, &y) != EOF){        if (x < 0 || y < 0){            break;        }        if (x == 0 || y == 0){            if (is_tree && single_root(range)){                printf("Case %d is a tree.\n", idx++);            }            else{                printf("Case %d is not a tree.\n", idx++);            }            is_tree = true;            range = 0;            make_set();            continue;        }        if (!is_tree){            continue;        }        range = x > range ? x : range;        range = y > range ? y : range;        flag[x] = flag[y] = true;        if (find_set(x) == find_set(y)){  //如果两者属于一个集合（也就是有共同祖先），并且两者还有父子关系，那么无法形成树，条件2            is_tree = false;        }        union_set(x, y);    }    return 0;}