In UDebug of this problem,there are one or more wrong input cases for which this code will show wrong output, but actually those input cases are invalid for this problem !! So don’t bother with the UDebug output, this code will be Accepted in UVa OJ since this code is for all valid input !! 😀
#include <bits/stdc++.h>
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define db double
#define ft float
#define ll long long
#define ull unsigned long long
#define ff first
#define ss second
#define vc vector
#define vi vector<int>
#define vll vector<long long>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define loop0(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define PI 3.14159265358979323846264338328
#define MAX 2004
using namespace std;
bool visited[MAX];
int dis[MAX];
vc <int> g[MAX];
map<ll,int> mp;
int bfs(int src,ll ttl)
{
int d=1;
dis[src]=0;
ms(visited,0);
ms(dis,0);
visited[src]=1;
queue<int> Q;
Q.push(src);
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int i=0; i<g[u].size(); i++)
{
int v=g[u][i];
if(!visited[v] && dis[u]+1<=ttl)
{
dis[v]=dis[u]+1;
d++;
visited[v]=1;
Q.push(v);
}
}
}
return d;
}
int main()
{
CIN;
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int edge;
for(int j=1;;)
{
cin>>edge;
if(edge==0) break;
int c=1;
for(int i=0; i<edge; i++)
{
ll from,to;
cin>>from>>to;
if(mp[from]==0)
{
mp[from]=c;
c++;
}
if(mp[to]==0)
{
mp[to]=c;
c++;
}
g[mp[from]].push_back(mp[to]);
g[mp[to]].push_back(mp[from]);
}
ll src,ttl;
while(cin>>src>>ttl)
{
if(src==0 && ttl==0) break;
int a=c-bfs(mp[src],ttl)-1;
cout<<"Case "<<j<<": "<<a<<" nodes not reachable from node "<<src<<" with TTL = "<<ttl<<"."<<endl;
j++;
}
loop0(i,MAX) g[i].clear();
mp.clear();
}
return 0;
}