Lowest common ancestor

1. Lowest Common Ancestor
Md. Shakil Ahmed
Software Engineer
Astha it research & consultancy ltd.
Dhaka, Bangladesh

2. lowest common ancestor (LCA)
• The lowest common ancestor (LCA) is a
concept in graph theory and computer
science. Let T be a rooted tree with n nodes.
The lowest common ancestor is defined
between two nodes v and w as the lowest
node in T that has both v and w as
descendants (where we allow a node to be a
descendant of itself).

3. lowest common ancestor (LCA)

4. LightOJ: 1101 - A Secret Mission
There will be N evil cities (numbered by 1, 2, ..., N) connected
by M bidirectional roads. There will be evil guards
patrolling the roads. Since they are not much intelligent,
the danger of travelling in each road is not the same. Alice
is going to travel from city s to city t. You can safely assume
that it's possible to travel from any city to another. John
the legend programmer has estimated the danger of each
road but since he is busy arranging contests, Alice is
depending on you now. Danger of a path from city s to city
t is defined as the maximum danger of any road on this
path. As you are one of the most talented programmers,
you will love to help Alice by finding the least dangerous
paths to prevent Evil Jack from doing harms to the Judges.

5. LightOJ: 1101 - A Secret Mission
Input
Input starts with an integer T (≤ 3), denoting the number of
test cases.
Each case contains two integers N, M (2 ≤ N ≤ 50000, 1 ≤ M ≤
105) denoting the number of cities and roads. Each of the
next M lines contains three integers: xi, yi, di (1 ≤ xi, yi ≤ N, xi
≠ yi, 1 ≤ di ≤ 104) - the cities connected by the ith road and its
degree of danger. Next line contains an integer q (1 ≤ q ≤
50000). Each of the next q lines contains two integers: si
and ti (1 ≤ si, ti ≤ N, si ≠ ti).
Output
For each test case, print the case number first. Then for each
query si ti, print the least dangerous path in a line.

6. LightOJ: 1101 - A Secret Mission

7. LightOJ: 1101 - A Secret Mission
int cmp( const void *a, const void
#include<stdio.h>
*b )
#include <stdlib.h> {
struct T T *p = (T *)a;
{ T *q = (T *)b;
int x;
return ( p->weight - q->weight );
int y; }
int weight;
}; int cmp1( const void *a, const void
*b )
{
struct TT
TT *p = (TT *)a;
{ TT *q = (TT *)b;
int x;
int y; return ( p->x - q->x );
int weight; }
};

8. LightOJ: 1101 - A Secret Mission
void dfs(int h1,int h2,int h3,int h4)
T a[100009]; {
TT A[100009]; PP[h1][0]=h2;
MX[h1][0]=h3;
int P[50009],ind[50009],N;
L[h1]=h4;
int Parent(long h1)
{ int h5;
if(P[h1]==-1)
return h1; for(h5=ind[h1];h5<N;h5++)
{
P[h1]=Parent(P[h1]);
if(A[h5].x!=h1)
return P[h1]; break;
} if(A[h5].y!=h2)
int PP[50009][20], MX[50009][20], dfs(A[h5].y,h1,A[h5].weight,h4+1);
L[50009];
int cas,cas1,n,m,i,h1,h2,j,flag,q; }
int q1,x,y,z,max; }

9. LightOJ: 1101 - A Secret Mission
int main() for(i=0;i<m;i++)
{ {
h1 = Parent(a[i].x);
scanf("%d",&cas); h2 = Parent(a[i].y);
if(h1!=h2)
for(cas1=1;cas1<=cas;cas1++)
{ {
scanf("%d %d",&n,&m); P[h1]=h2;
A[N].x = a[i].x;
for(i=0;i<m;i++) A[N].y = a[i].y;
scanf("%d %d %d",&a[i].x,&a[i].y,&a[i].weight);
A[N].weight = a[i].weight;
qsort(a,m,sizeof(T),cmp); N++;
A[N].x = a[i].y;
for(i=1;i<=n;i++) A[N].y = a[i].x;
P[i]=-1; A[N].weight = a[i].weight;
N = 0; N++;
}
}

10. LightOJ: 1101 - A Secret Mission
qsort(A,N,sizeof(TT),cmp1); for(i=1;i<=n;i++)
{
for(i=1;i<=n;i++) if(PP[i][j-1]!=-1 && PP[PP[i][j-1]][j-1]!=-1)
{ {
ind[i]=-1; PP[i][j]= PP[PP[i][j-1]][j-1];
} MX[i][j]=MX[i][j-1];
if(MX[i][j]<MX[PP[i][j-1]][j-1])
for(i=0;i<N;i++)
MX[i][j]=MX[PP[i][j-1]][j-1];
if(ind[A[i].x]==-1) flag = 1;
ind[A[i].x] = i; }
else
dfs(1,-1,-1,0); PP[i][j]=-1;
}
for(j=1;;j++)
{ if(flag==0)
flag = 0; break;
}

11. LightOJ: 1101 - A Secret Mission
scanf("%d",&q); while(L[x]!=L[y])
printf("Case %d:n",cas1); {
for(q1=1;q1<=q;q1++) for(i=0;;i++)
{ {
scanf("%d %d",&x,&y); if(PP[y][i]==-1||L[PP[y][i]]<L[x])
max = 0; {
y = PP[y][i-1];
if(L[x]>L[y]) break;
{ }
z = x; else if(MX[y][i]>max)
x = y; max = MX[y][i];
y = z; }
}
}

12. LightOJ: 1101 - A Secret Mission
while(x!=y) else
{ {
for(i=0;i<20;i++) x = PP[x][i-1];
y = PP[y][i-1];
{ }
if(PP[x][i]==PP[y][i]) break;
{ }
if(i==0) else
{
{
if(MX[x][i]>max)
if(MX[x][i]>max) max = MX[x][i];
max = MX[x][i]; if(MX[y][i]>max)
if(MX[y][i]>max) max = MX[y][i];
max = MX[y][i]; }} }
printf("%dn",max);
x = PP[x][i];
} }
y = PP[y][i]; return 0;
} }

13. Sample
• LightOJ:
http://www.lightoj.com/volume_problemcate
gory.php?user_id=3060&category=Range
%20Minimum%20Query/Lowest%20Common
%20Ancestor

14. Thanks!