ACM1325Is it A tree?
通过这道简单而又坑人的题目,练习并查集和set 容器的使用;
Is It A Tree?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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.
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
Case 2 is a tree.
Case 3 is not a tree.
#include<iostream>
#include<cstring>
#include<set>
using namespace std;
const int N=;
int p[N];
bool flag;
int find(int x)
{
if(x==p[x])
return x;
return p[x]=find(p[x]);
}
void link(int a,int b)
{
int fa=find(a);
int fb=find(b);
if(fa==fb)
flag=true;
else
{
if(fb!=b)
flag=true;
p[fb]=fa;
}
}
int main()
{
int n,m;
int cd=;
while(true)
{
set<int>q;
flag=false;
for(int i=;i<N;i++)
p[i]=i;
scanf("%d %d",&n,&m);
if(n==&&m==)
{
printf("Case %d is a tree.\n",cd);
cd++;
continue;
}
if(n<&&m<)break;
q.insert(n);
q.insert(m);
link(n,m);
while(scanf("%d%d",&n,&m)==&&n&&m)
{
q.insert(n);
q.insert(m);
link(n,m);
}
int count=-;
set<int>::iterator it;
for(it=q.begin();it!=q.end();it++)
{
if(p[*it]==*it)
count++;
}
if(count>)flag=true;
if(flag==true)
printf("Case %d is not a tree.\n",cd);
else printf("Case %d is a tree.\n",cd);
cd++;
}
return ;
}
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