HDU 1325（并查集）

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23312    Accepted Submission(s): 5338

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.

 

Source

North Central North America 1997

 

Recommend

Ignatius.L   |   We have carefully selected several similar problems for you:  1856 1232 1233 1102 1875

/*
HDU 1272的升级版，建议先做1272.
 有向图，判断是不是树。
注意点：
1 2 2 1 0 0 No
1 2 1 2 0 0 No
1 2 3 4 0 0 No
0 0 Yes
ru[i]==0的个数要等于1.
*/
#include<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + ;
int ru[maxn], a[maxn];
int main ()
{
    int ic = ;
    int u, v;
    bool fg = true;
    while(fg)
    {
        memset(ru, , sizeof(ru));
        bool ans = true;
        map<int, int> mp;
        int nu = ;
        while(true)
        {
            scanf("%d%d", &u, &v);
            if(u ==  && v == )
            {
                if(ans)
                {
                    int t = ;
                    for(int i = ; i <= nu; i++)
                    {
                        if(ru[a[i]] <= )
                        {
                            if(ru[a[i]] == )
                            {
                                t++;
                                if(t > )
                                {
                                    ans = false;
                                    break;
                                }
                            }
                        }
                        else
                        {
                            ans = false;
                            break;
                        }
                    }
                }
                if(ans)
                    printf("Case %d is a tree.\n", ++ic);
                else
                    printf("Case %d is not a tree.\n", ++ic);
                break;
            }
            else if(u <  && v < )
            {
                fg = false;
                break;
            }
            if(!mp.count(u))
            {
                a[++nu] = u;
                mp[u] = ;
            }
            if(!mp.count(v))
            {
                a[++nu] = v;
                mp[v] = ;
            }
            if(ans)
            {
                ru[v]++;
                int t = u *  + v;
                if(mp.count(t))
                    ans = false;
                else
                    mp[t] = ;
                 t = u + v * ;
                if(mp.count(t))
                    ans = false;
                else
                    mp[t] = ;
            }
        }
    }
    return ;
}