【HDU - 1325】Is It A Tree?

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Is It A Tree?

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

Total Submission(s): 26113    Accepted Submission(s): 5966


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 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.

//代码如下：

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

const int maxn = 1e6 + 10;

int par[maxn];
int visit[maxn]; //记录点是否出现
int num[maxn]; //记录点的入度 如果大于2 就成环了 可以在纸上花一下 就肯定不行了

void init()
{
for (int i = 0; i <= maxn; i++)
par[i] = i, num[i] = 0, visit[i] = 0;
}

int find(int x)
{
return x == par[x] ? x : par[x] = find(par[x]);
}

void unite(int a, int b)
{
int fa = find(a);
int fb = find(b);
if (fa != fb)
par[fa] = fb;
}

int main()
{
int n, m;
int t = 1;
while (scanf("%d%d", &n, &m) != EOF)
{
if (n < 0 && m < 0)
break;
init();
if (m == 0 && m == 0)
{
printf("Case %d is a tree.\n", t++);
continue;
}
bool flag = true;
while (n || m)
{
visit
= 1, visit[m] = 1;
num[m]++;
if (num[m] > 1) //入度大于1
flag = false;
if (find(n) == find(m)) //说明这两个点已经在一棵树上了 再加入的话就会出现成环或者重边 判断两个点在不在一棵树上一可以用这种方法
flag = false;
unite(n, m);
scanf("%d%d", &n, &m);
}
int ans = 0, sum = 0; //ans记录出现的点的个数 下面防止单独的一个点 sum标记集合的个数 防止森林的出现
//下面开始特判 这个题坑点比较多
for (int i = 0; i <= maxn; i++)
{
if (par[i] == i && visit[i] == 1)
sum++;
if (visit[i] == 1)
ans++;
}
if (sum > 1)
flag = false;
if (sum == 1 && ans == 1) //这个就是上面说的单独的一个点
flag = false;
if (flag)
printf("Case %d is a tree.\n", t++);
else
printf("Case %d is not a tree.\n", t++);
}
return 0;
}