zoukankan      html  css  js  c++  java
  • POJ1308——Is It A Tree?

    Is It A Tree?

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 22631   Accepted: 7756

    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.

    Source


    推断一些点形成的结构是否是一棵树。坑点好多,用并查集推断连通支数以及环,推断每一个点的入度。推断是否有自己连向自己的边


    #include <map>
    #include <set>
    #include <list>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    int father[100010];
    bool vis[100010];
    int in_deg[100010];
    
    int find(int x)
    {
        if (father[x] == -1)
        {
            return x;
        }
        return father[x] = find(father[x]);
    }
    
    void init()
    {
        memset( father, -1, sizeof(father));
        memset( vis, 0, sizeof(vis));
        memset(in_deg, 0, sizeof(in_deg));
    }
    
    int main()
    {
        int x, y;
        int icase = 1;
        while (~scanf("%d%d", &x, &y))
        {
            if (x == -1 && y == -1)
            {
                break;
            }
            if (x == 0 && y == 0)
            {
                printf("Case %d is a tree.
    ", icase++);
                continue;
            }
            init();
            map<int, int> bianhao;
            bianhao.clear();
            int cnt = 0;
            bool flag = true;
            while (x && y)
            {
                if (!x && !y)
                {
                    break;
                }
                if (x == y)
                {
                    flag = false;
                }
                if (flag && !vis[x])
                {
                    vis[x] = 1;
                    bianhao[x] = ++cnt;
                }
                if (flag && !vis[y])
                {
                    vis[y] = 1;
                    bianhao[y] = ++cnt;
                }
                if (flag)
                {
                    int a = find(bianhao[x]);
                    int b = find(bianhao[y]);
                    if (a == b)
                    {
                        flag = false;
                    }
                    father[a] = b;
                    in_deg[bianhao[y]]++;
                }
                scanf("%d%d", &x, &y);
            }
            if (!flag)
            {
                printf("Case %d is not a tree.
    ", icase++);
                continue;
            }
            int ans = 0;
            // for (int i = 1; i <= cnt; ++i)
            // {
                // printf("%d ", in_deg[i]);
            // }
            for (int i = 1; i <= cnt; ++i)
            {
                if (in_deg[i] == 0)
                {
                    ans++;
                }
                if (ans >= 2)
                {
                    break;
                }
                if (in_deg[i] >= 2)
                {
                    flag = false;
                    break;
                }
            }
            if (ans >= 2 || !flag)
            {
                printf("Case %d is not a tree.
    ", icase++);
                continue;
            }
            printf("Case %d is a tree.
    ", icase++);
        }
        return 0;
    }




  • 相关阅读:
    为ccflow增加禁用用户立刻生效功能
    关于工作流引擎授权问题的需求变更
    sql server 2005 安装过程中出现错误Insatalling performance countter: Cannot create a file when that file already exists.
    web.config中错误
    三个SQL视图查出所有SQL Server数据库字典
    恢复备份的数据库
    sql语句读取excel数据
    It is an error to use a section registered allowDefinition='MachineToApplication' beyond application level. 错误
    DOS命令实现创建文件夹
    如何查看sql server 的版本(网摘)
  • 原文地址:https://www.cnblogs.com/liguangsunls/p/7001021.html
Copyright © 2011-2022 走看看