POJ1308——Is It A Tree?

标签：并查集

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

Output

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

判断一些点形成的结构是否是一棵树，坑点好多，用并查集判断连通支数以及环，判断每个点的入度，判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.\n", 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.\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;
}

POJ1308——Is It A Tree?

标签：并查集

原文地址：http://blog.csdn.net/guard_mine/article/details/41318859