Problem Link : http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1879
Solution Idea: Find the LCA. Then find the middle node and middle+1 node on the path connection two nodes which are given in the query. Here I use an extra DFS to find the path on a reverse graph g2.
#include <bits/stdc++.h> #define pii pair <int,int> #define pll pair <long long,long long> #define sc scanf #define pf printf #define Pi 2*acos(0.0) #define ms(a,b) memset(a, b, sizeof(a)) #define pb(a) push_back(a) #define MP make_pair #define db double #define ll long long #define EPS 10E-10 #define ff first #define ss second #define sqr(x) (x)*(x) #define D(x) cout<<#x " = "<<(x)<<endl #define VI vector <int> #define DBG pf("Hi\n") #define MOD 1000000007 #define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0) #define SZ(a) (int)a.size() #define sf(a) scanf("%d",&a) #define sfl(a) scanf("%lld",&a) #define sff(a,b) scanf("%d %d",&a,&b) #define sffl(a,b) scanf("%lld %lld",&a,&b) #define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c) #define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c) #define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++) #define loop(i,n) for(int i=0;i<n;i++) #define loop1(i,n) for(int i=1;i<=n;i++) #define REP(i,a,b) for(int i=a;i<b;i++) #define RREP(i,a,b) for(int i=a;i>=b;i--) #define TEST_CASE(t) for(int z=1;z<=t;z++) #define PRINT_CASE printf("Case %d: ",z) #define LINE_PRINT_CASE printf("Case %d:\n",z) #define CASE_PRINT cout<<"Case "<<z<<": " #define all(a) a.begin(),a.end() #define intlim 2147483648 #define infinity (1<<28) #define ull unsigned long long #define gcd(a, b) __gcd(a, b) #define lcm(a, b) ((a)*((b)/gcd(a,b))) using namespace std; /*----------------------Graph Moves----------------*/ //const int fx[]={+1,-1,+0,+0}; //const int fy[]={+0,+0,+1,-1}; //const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move //const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move //const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move //const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move /*------------------------------------------------*/ /*-----------------------Bitmask------------------*/ //int Set(int N,int pos){return N=N | (1<<pos);} //int reset(int N,int pos){return N= N & ~(1<<pos);} //bool check(int N,int pos){return (bool)(N & (1<<pos));} /*------------------------------------------------*/ #define mx 5005 int n,q; vector<int>g1[mx],g2[mx]; int sparse_par[mx][14],sparse_dis[mx][14]; int level[mx],par[mx]; void dfs(int u, int cnt, int from) { par[u]=from; level[u]=cnt; g2[u].pb(from); for(int i=0;i<SZ(g1[u]);i++) { int v=g1[u][i]; if(v!=from) { dfs(v,cnt+1,u); } } } void build_talbe() { for(int i=1;i<=n;i++) { sparse_par[i][0]=par[i]; sparse_dis[i][0]=1; } sparse_dis[1][0]=0; for(int j=1; 1<<j<=n;j++) { for(int i=1;i<=n;i++) { sparse_par[i][j]=sparse_par[sparse_par[i][j-1]][j-1]; sparse_dis[i][j]=sparse_dis[i][j-1]+sparse_dis[sparse_par[i][j-1]][j-1]; } } } int LCA; int query(int p, int q) { if(level[p]<=level[q]) swap(p,q); int log=log2(level[p]); int ret=0; for(int i=log;i>=0;i--) { if(level[p]-(1<<i)>=level[q]) { ret+=sparse_dis[p][i]; p=sparse_par[p][i]; } } if(p==q){LCA=p; return ret;} for(int i=log;i>=0;i--) { if(sparse_par[p][i]!=sparse_par[q][i]) { ret+=sparse_dis[p][i]; ret+=sparse_dis[q][i]; p=sparse_par[p][i]; q=sparse_par[q][i]; } } ret+=2; LCA=par[p]; return ret; } vector<int>temp; int dfs1(int u, int cnt) { if(u==LCA) return u; temp.pb(u); if(cnt==0) return u; for(int i=0;i<SZ(g2[u]);i++) return dfs1(g2[u][i],cnt-1); } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); while(sf(n) && n) { loop(i,n-1) { int a,b; sff(a,b); g1[a].pb(b); g1[b].pb(a); } dfs(1,0,1); build_talbe(); sf(q); while(q--) { int a,b; sff(a,b); int dis=query(a,b)+1; if(dis%2==0) { temp.clear(); dfs1(a,n); temp.pb(LCA); int x=SZ(temp); dfs1(b,n); reverse(temp.begin()+x,temp.end()); int aa=temp[(dis/2)-1]; int bb=temp[dis/2]; pf("The fleas jump forever between %d and %d.\n",min(aa,bb),max(aa,bb)); } else { int p; if(level[a]>level[b]) p=a; else p=b; pf("The fleas meet at %d.\n",dfs1(p,dis/2)); } } for(int i=0;i<=n;i++) { g1[i].clear(); g2[i].clear(); } } return 0; }