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在hdu 1272 的基础上稍加修改就ac了
1272已经判断了无向情况下是否是树形结构,因此我们只需要多判断一下入度为0的点是否只有一个就好了
#include<iostream>
#define maxn 100000+5
using namespace std;
int a,b;
int flag;
int father[maxn];
int sign[maxn];
int r[maxn];
void ready()
{
for(int i=1;i<maxn;i++) r[i]=0,father[i]=i,sign[i]=0;
}
int dfs(int x)
{
if(father[x]!=x)
{
father[x]=father[father[x]];
}
return father[x];
}
void build(int x,int y)
{
if(x==y) flag=0;
else
{
if(dfs(x)==dfs(y)) flag=0;
else
{
father[dfs(x)]=dfs(y);
}
}
r[y]++;
}
int main()
{
int casee=1;
while(cin>>a>>b)
{
ready();
flag=1;
if(a<0&&b<0){break;}
if(!a&&!b) {cout<<"Case "<<casee++<<" is a tree."<<endl;continue;}
build(a,b);sign[a]=sign[b]=1;
while(cin>>a>>b&&a&&b)
{
build(a,b);
sign[a]=sign[b]=1;
}
int k=0,d=0;
for(int i=1;i<maxn;i++)
{
if(sign[i]&&dfs(i)==i) k++;
if(sign[i]&&r[i]==0) d++;
}
if(k==1&&flag&&d==1) cout<<"Case "<<casee++<<" is a tree."<<endl;
else cout<<"Case "<<casee++<<" is not a tree."<<endl;
}
return 0;
}
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原文地址:http://blog.csdn.net/zafkiel_nightmare/article/details/45512133