10000 - Longest Paths

[alu_mathics]

very complex implementation of BFS indeed.
you have used adjacency list.
while(v && e)
{
j=0;
while(p[v].paths[j]>=0)j++;
p[v].paths[j]=e;
scanf("%d %d",&v,&e);
}
for the colored part you can use another counter array.it will reduce the time.


set max=-1;


for(j=0;j<=N;j++)
{
if(p[j].dis>max)
{
max=p[j].dis;
end=j;
}
}


why j=0,don't know.

no need of the condition , atleast for this program
if(counter>p[p[q].paths[j]].dis)
so u can remove it.

by the way the matrix method implementation is easier for this program.

[Ali Arman Tamal]

I'm getting TLE don't know why...

Here is my code

Code: Select all

#include <stdio.h>

int a[101][101],parent[101],d[101];
int n,root,casecount=0;

void process(int k);
void findpath();

int main()
{

	int i,j;

	while(1)
	{
	scanf("%d",&n);
	if(n==0)break;

	casecount++;

	for(i=1;i<=n;i++)
	{
		parent[i]=-1;
		d[i]=0;
		for(j=1;j<=n;j++)a[i][j]=0;
	}

	scanf("%d",&root);
	parent[root]=-1;

	while(1)
	{
	scanf("%d %d",&i,&j);
	if(i==0 && j==0)break;
	a[i][j]=1;
	}

	process(root);

	findpath();

	}

return 1;
}



void process(int k)
{

	int i=1;

	while(i<=n)
	{
		if(a[k][i]==1)if(d[k]+1>d[i]){ d[i]=d[k]+1; parent[i]=k; }
		i++;
	}

	i=1;
	while(i<=n){ if(a[k][i]==1)process(i); i++; }

}


void findpath()
{
int k,max=root,count;

for(k=1;k<=n;k++)
{
if(d[k]>d[max])max=k;
else if(d[k]==d[max] && k<max)max=k;
}

count=0;

if(max!=root){
	k=max;
	while(parent[k]!=-1){ k=parent[k]; count++; }
}

printf("Case %d: The longest path from %d has length %d, finishing at %d.\n\n",casecount,root,count,max);

}

The problem description says

You can assume that the graph has no cycles (there is no path from any node to itself),

My code will certainly give TLE if the graph is cyclic, but I'm not sure if it is the reason.


THANKS FOR READING !! [/quote]

[mohiul alam prince]

hi
i have used only BFS algorithm and just got AC
u can use BFS algorithm

[Ali Arman Tamal]

The previous attempt was buggy. Now I got AC with BFS

[ImLazy]

Does the pair of p and q means:
1.There is a path from p to q. And there may be some other nodes between p and q. For example, it can be p->1->2->q
Or
2.There is a edge from p to q in the graph. That is p->q, no other nodes between them.
Which of the two understanding is right?

[CDiMa]

Does the pair of p and q means:
1.There is a path from p to q. And there may be some other nodes between p and q. For example, it can be p->1->2->q
Or
2.There is a edge from p to q in the graph. That is p->q, no other nodes between them.
Which of the two understanding is right?

The latter will do...

Ciao!!!

Claudio

[ImLazy]

Thank you, CDiMa.
And I was so silly because the first understanding doesn't fit the sample output.

[kruslan]

Cant find the problem... I'm always getting WA...

Code: Select all

const maxn = 100;

var a : array [0..maxn,0..maxn] of integer;
    q,l : array [0..maxn*maxn] of integer;
    c,n,i,x,y,st,max,maxi,t,j,yk,yp,cur : integer;
    w : array [0..maxn+1,0..maxn+1] of boolean;
    h : boolean;

begin
{  assign(input,'input.txt'); reset(input);
  assign(output,'output.txt'); rewrite(output);
}
  n:=1;
  c:=0;
  while n<>0 do
  begin
    readln(n);
    if n>0 then
    begin
      {}
      readln(st);
      fillchar(a,sizeof(a),0);
      x:=1; y:=1;
      while not((x=0)and(y=0)) do
      begin
        readln(x,y);
        if not((x=0)and(y=0)) then
        begin
          inc(a[x,0]);
          a[x,a[x,0]]:=y;
        end;
      end;
      {}
      max:=0; maxi:=st;
      yk:=1; yp:=1; q[yk]:=st; l[yk]:=0;
      while yk<=yp do
      begin
        cur:=q[yk];
        for i:=1 to a[cur,0] do
        begin
          if w[a[cur,i],l[yk]+1]=false then
          begin
            w[a[cur,i],l[yk]+1]:=true;
            inc(yp);
            q[yp]:=a[cur,i];
            l[yp]:=l[yk]+1;
          end;
        end;
        if (l[yk]>max)or((l[yk]=max)and(q[yk]<maxi)) then
        begin
          max:=l[yk];
          maxi:=q[yk];
        end;
        inc(yk);
      end;
      {}
      inc(c);
      writeln('Case ',c,': The longest path from ',st,' has length ',max,', finishing at ',maxi,'.');
      writeln;
      {}
    end;
  end;
{  close(input); close(output);}
end.

[N|N0]

I tested it on:

Code: Select all

2
1
1 2
0 0
5
3
1 2
3 5
3 1
2 4
4 5
0 0
5
5
5 1
5 2
5 3
5 4
4 1
4 2
0 0
2
1
1 2
0 0
0

And it produced:

Code: Select all

Case 1: The longest path from 1 has length 1, finishing at 2.

Case 2: The longest path from 3 has length 4, finishing at 5.

Case 3: The longest path from 5 has length 2, finishing at 1.

Case 4: The longest path from 1 has length 0, finishing at 1.

The output should be:

Code: Select all

Case 1: The longest path from 1 has length 1, finishing at 2.

Case 2: The longest path from 3 has length 4, finishing at 5.

Case 3: The longest path from 5 has length 2, finishing at 1.

Case 4: The longest path from 1 has length 1, finishing at 2.

... since input data for Case 1 and Case 2 is the same.

Try initializing the w[][] variable before anything else is done.

Code: Select all

for i := 0 to maxn+1 do
  for j := 0 to maxn+1 do
    w[i][j] := false;

[kruslan]

What a silly mistake....
Thank you very much

I added line fillchar(w,sizeof(w),false);
And got AC

[pipo]

hi all...

I solved this problem(10000-longest paths) using back-tracking algorithm technic...

it gave me correct answers for the sample input..

but.. judger replied the code was TLE...

it is so fool of me...

how do I speed up using back-tracking algorithm technic ??

and would you give me a sample test case which occurs TLE ??

thanks in advance...

ps.. sorry for poor english... excuse me

[arif_pasha]

You can speed it up simply using some pruning. say you are in node n and you know it was previously explored in depth d[n]. if you reach this node again from another path with depth curr_depth you only explore it if d[n]<curr_depth.

hope it helps..

[chunyi81]

Using DFS should work as well. I got AC using DFS and C++ STL for a reasonable time. Keeping track of visited nodes will help.

[898989]

Salamo Aleko
As for ur code u need to prune ur search as guided above........
But you also can solve this problem easily by using floyd warshal algorithm to find shortest path... Only you need to minor changes
To make it find the longest path

Using bfs or dfs I got ac in 0.387s
Using floyd I got ac in 1.275

http://acm.uva.es/problemset/usersjudge.php?user=8957CA

[smilitude]

Is my algo wrong??

Code: Select all

/*
 * 10000 longest path
 * submission 1 WA
 * coded at 1157am , may 11, 2006
 */

#include <cstdio>
#include <queue>
#define max(a,b) (((a)>(b)) ? (a):(b))
using namespace std;

int main() {
    struct {
        int edge[100];
        int nedge;
        int dist;
        bool processed;
        bool discovered;
    }node[101];
    
    int n,s,i,j,a,b,cur,max,end,cases=0;
    int cedge;
    queue<int> Q;
    
    while(scanf("%d",&n)==1 && n) {
        scanf("%d",&s);
        
        //init
        for(i=1;i<=n;i++) {
            node[i].nedge=0;
            node[i].dist=0;
            node[i].processed=node[i].discovered=false;
        }
        
        //input
        while(true) {
            scanf("%d%d",&a,&b);
            if(a==0 && b==0) break;
            node[a].edge[node[a].nedge++]=b;
        }
        
        //bfs
        Q.push(s);
        while(!Q.empty()) {
            cur=Q.front();
            Q.pop();
            for(i=0;i<node[cur].nedge;i++) {
                cedge=node[cur].edge[i];
                node[cedge].dist=max(node[cedge].dist,node[cur].dist+1);
                if(node[cedge].processed) continue;
                if(!node[cedge].discovered) {
                    node[cedge].discovered=true;
                    Q.push(cedge);
                }
            }
            node[cur].processed=true;
        }
        
        //finding max
        max=0;
        for(i=1;i<=n;i++) {
            if(node[i].dist>max) {
                max=node[i].dist;
                end=i;
            }
        }
        printf("Case %d: The longest path from %d has length %d, finishing at %d.\n\n",++cases,s,max,end);
    }
return 0;
}

I am getting wrong answer,
Can anyone help me ?