程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ1308¡ª¡ªIs It A Tree?

POJ1308¡ª¡ªIs It A Tree?

編輯:C++入門知識

POJ1308¡ª¡ªIs 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 Z†·Ÿ"http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">vciBub3QuCgo8cCBjbGFzcz0="pst">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


ÅжÏһЩµãÐγɵĽṹÊÇ·ñÊÇÒ»¿ÃÊ÷£¬¿ÓµãºÃ¶à£¬Óò¢²é¼¯ÅжÏÁ¬Í¨Ö§ÊýÒÔ¼°»·£¬ÅжÏÿ¸öµãµÄÈë¶È£¬ÅжÏÊÇ·ñÓÐ×Ô¼ºÁ¬Ïò×Ô¼ºµÄ±ß


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

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.\n", icase++);
            continue;
        }
        init();
        map 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.\n", 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.\n", icase++);
            continue;
        }
        printf("Case %d is a tree.\n", icase++);
    }
    return 0;
}




  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved