【并查集】Is it a tree？

最新推荐文章于 2020-09-10 10:51:59 发布

Whjpji

最新推荐文章于 2020-09-10 10:51:59 发布

阅读量626

点赞数

CC 4.0 BY-SA版权

分类专栏： OI 文章标签： tree integer each collections pair structure

本文链接：https://blog.youkuaiyun.com/Whjpji/article/details/7470033

OI 专栏收录该内容

177 篇文章

订阅专栏

本文深入探讨了并查集的基本概念及其在判断树形结构有效性中的应用，通过实例展示了如何通过并查集来判断一组节点连接是否形成合法的树形结构。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

很裸的并查集，注意森林不是树。
Accode：

#include <cstdio>
#include <string>
#include <set>

const int maxN = 1010;
std::set <int> S; int F[maxN];
std::set <int>::iterator iter;

inline int getint()
{
    int res = 0; char tmp; bool sgn = 1;
    do tmp = getchar();
    while (!isdigit(tmp) && tmp - '-');
    if (tmp == '-') {sgn = 0; tmp = getchar();}
    do res = (res << 3) + (res << 1) + tmp - '0';
    while (isdigit(tmp = getchar()));
    return sgn ? res : -res;
}

int Find(int x)
{return F[x] == x ? x : F[x] = Find(F[x]);}

int main()
{
    freopen("tree.in", "r", stdin);
    freopen("tree.out", "w", stdout);
    for (int T = 1; ; ++T)
    {
        for (int i = 1; i < maxN; ++i) F[i] = i;
        S.clear(); int x, y; bool flag = 1;
        while ((x = getint(), y = getint()) > 0)
        {
            if (S.find(x) == S.end()) S.insert(x);
            if (S.find(y) == S.end()) S.insert(y);
            if (Find(x) == Find(y)) flag = 0;
            else F[F[y]] = F[x];
        }
        if (x < 0 || y < 0) break;
        if (flag)
        {
            iter = S.begin();
            for (; iter != S.end(); ++iter)
            if (Find(*iter) - Find(*(S.begin())))
            {flag = 0; break;}
        } //要注意空树，这里很容易出错。
        printf("Case %d is%s a tree.\n",
               T, flag ? "" : " not");
    }
    return 0;
}