280  Vertex
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Ok, let's make a simple example:Ferdous wrote:Don't I need to search for every query? I can't understand what you have told. Would you please explain a little more?
Code: Select all
4
1 2 0
2 3 0
3 4 0
0
2 2 1
0
Code: Select all
1>2>3>4
When you process the second query you start from 1 and visit 2,3 and 4. But you already visited the graph starting from 2 earlier. You can add nodes reachable from 2 to the set of nodes reachable from 1 without revisiting the entire graph.
In my program after getting input I generate the sets of unreachable nodes for all nodes in the graph and then simply output the result for every node queried.
Ciao!!!
Claudio

 Experienced poster
 Posts: 115
 Joined: Tue Apr 06, 2004 7:04 pm
 Location: Venezuela
280  Vertex
Hi !! to everyone I get WA, this is my algo, its something i miss ?? please Help me
1. DO Floyd  warshall
2. count the inaccessible vertex and put in a array
3. print the count of inaccessible vertex
4. print the numbers of inaccessible vertex from the array
any I/O will be usefull, or any hints please help
this is some I/O is right??
Input:
ouput
Thanks in advance
Keep posting !!
1. DO Floyd  warshall
2. count the inaccessible vertex and put in a array
3. print the count of inaccessible vertex
4. print the numbers of inaccessible vertex from the array
any I/O will be usefull, or any hints please help
this is some I/O is right??
Input:
Code: Select all
3
1 2 0
2 2 0
3 1 2 0
0
2 1 2
4
1 2 0
2 3 0
3 4 0
0
2 2 1
5
1 2 0
2 3 0
3 4 0
4 5 0
1 1 0
0
1 1
0
Code: Select all
2 1 3
2 1 3
2 1 2
1 1
0
Keep posting !!
Re: 280 Vertex WA Help!!
My AC program gives the same output.Ghust_omega wrote: any I/O will be usefull, or any hints please help
this is some I/O is right??
[...]
Try this input :
Code: Select all
7
1 2 0
2 3 4 0
3 1 0
4 5 0
5 4 0
6 7 0
7 6 0
0
7 1 2 3 4 5 6 7
0
Code: Select all
2 6 7
2 6 7
2 6 7
5 1 2 3 6 7
5 1 2 3 6 7
5 1 2 3 4 5
5 1 2 3 4 5
Claudio

 Experienced poster
 Posts: 115
 Joined: Tue Apr 06, 2004 7:04 pm
 Location: Venezuela

 Experienced poster
 Posts: 154
 Joined: Sat Apr 17, 2004 9:34 am
 Location: EEE, BUET
You may try this input
Meanwhile, what about using DFS?
Code: Select all
5
0
5 1 2 3 4 5
4
1 2 3 4 0
2 3 4 0
3 4 0
0
4 1 2 3 4
You should never take more than you give in the circle of life.

 Experienced poster
 Posts: 115
 Joined: Tue Apr 06, 2004 7:04 pm
 Location: Venezuela
Hi !! Mohammad Mahmudur Rahman, thanks for the I/O This is the ouput:
I think that floyd solve the problem too, any suggestions??
Thanks in advance
Keep posting !!
Code: Select all
5 1 2 3 4 5
5 1 2 3 4 5
5 1 2 3 4 5
5 1 2 3 4 5
5 1 2 3 4 5
1 1
2 1 2
3 1 2 3
4 1 2 3 4
Thanks in advance
Keep posting !!

 Experienced poster
 Posts: 115
 Joined: Tue Apr 06, 2004 7:04 pm
 Location: Venezuela

 Experienced poster
 Posts: 154
 Joined: Sat Apr 17, 2004 9:34 am
 Location: EEE, BUET
keep getting WA.....
I don't see what can be possibly wrong with my code, but keep getting WA. I'm using Floyd  Warshall.
[cpp]#include <cstdio>
#include <string>
using namespace std;
char data[201][201];
int ans[201];
int n, m, a, b, cnt;
int main(){
while(scanf("%d", &n) == 1 && n){
memset(data, 0, sizeof(data));
while(scanf("%d", &a) == 1 && a)
while(scanf("%d", &b) == 1 && b){
data[a] = 1;
a ^= b ^= a ^= b;
}
for(int k = 1; k <= n; k++)
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
data[j] = (data[k] && data[k][j]);
scanf("%d", &m);
while(m && scanf("%d", &a) == 1){
cnt = 0;
for(int i = 1; i <= n; i++)
if(!data[a]) ans[cnt++] = i;
printf("%d", cnt);
for(int i = 0; i < cnt; i++) printf(" %d", ans);
printf("\n");
}
}
return 0;
}
[/cpp]
[cpp]#include <cstdio>
#include <string>
using namespace std;
char data[201][201];
int ans[201];
int n, m, a, b, cnt;
int main(){
while(scanf("%d", &n) == 1 && n){
memset(data, 0, sizeof(data));
while(scanf("%d", &a) == 1 && a)
while(scanf("%d", &b) == 1 && b){
data[a] = 1;
a ^= b ^= a ^= b;
}
for(int k = 1; k <= n; k++)
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
data[j] = (data[k] && data[k][j]);
scanf("%d", &m);
while(m && scanf("%d", &a) == 1){
cnt = 0;
for(int i = 1; i <= n; i++)
if(!data[a]) ans[cnt++] = i;
printf("%d", cnt);
for(int i = 0; i < cnt; i++) printf(" %d", ans);
printf("\n");
}
}
return 0;
}
[/cpp]

 Experienced poster
 Posts: 115
 Joined: Tue Apr 06, 2004 7:04 pm
 Location: Venezuela

 Experienced poster
 Posts: 115
 Joined: Tue Apr 06, 2004 7:04 pm
 Location: Venezuela
Hi !! minskcity this is the part I change
[c]
while(scanf("%d", &n) == 1 && n){
memset(data, 0, sizeof(data));
while(scanf("%d", &a) == 1 && a)
while(scanf("%d", &b) == 1 && b){
..............
}
for( k = 1; k <= n; k++)
for( i = 1; i <= n; i++)
for( j = 1; j <= n; j++)
...............
[/c]
Hope it Helps
Keep posting
[c]
while(scanf("%d", &n) == 1 && n){
memset(data, 0, sizeof(data));
while(scanf("%d", &a) == 1 && a)
while(scanf("%d", &b) == 1 && b){
..............
}
for( k = 1; k <= n; k++)
for( i = 1; i <= n; i++)
for( j = 1; j <= n; j++)
...............
[/c]
Hope it Helps
Keep posting
I used just dfs & cant find out why WA?
here is my code
#include<stdio.h>
#include<malloc.h>
#define SIZE 200
typedef struct node{
int paths[SIZE];
int isVisited;
}vertex;
vertex *p;
int reachable;
void DFS(int i)
{
int j;
j=0;
while(p.paths[j]>=0 && p[p.paths[j]].isVisited==1)
{
p[p.paths[j]].isVisited=1;
reachable++;
DFS(p.paths[j]);
j++;
}
}
void doDFS(int N,int start)
{
int i;
DFS(start);
printf("%d",Nreachable);
for(i=1;i<=N;i++)
{
if(p.isVisited==1)printf(" %d",i);
}
printf("\n");
}
void refresh()
{
int i,j;
for(i=0;i<SIZE;i++)
{
for(j=0;j<SIZE;j++)
{
p.paths[j]=1;
}
p.isVisited=1;
}
}
void main()
{
int N,i,e,j;
p=(vertex*)malloc(SIZE*sizeof(vertex));
while(scanf("%d",&N)==1 && N)
{
refresh();
while(scanf("%d",&i) ==1 && i)
{
while(scanf("%d",&e)==1 && e)
{
j=0;
while(p.paths[j]>=0)j++;
p.paths[j]=e;
}
}
scanf("%d",&i);
while(i>0)
{
scanf("%d",&j);
reachable=0;
doDFS(N,j);
for(e=0;e<=N;e++)
{
p[e].isVisited=1;
}
}
}
free(p);
}
here is my code
#include<stdio.h>
#include<malloc.h>
#define SIZE 200
typedef struct node{
int paths[SIZE];
int isVisited;
}vertex;
vertex *p;
int reachable;
void DFS(int i)
{
int j;
j=0;
while(p.paths[j]>=0 && p[p.paths[j]].isVisited==1)
{
p[p.paths[j]].isVisited=1;
reachable++;
DFS(p.paths[j]);
j++;
}
}
void doDFS(int N,int start)
{
int i;
DFS(start);
printf("%d",Nreachable);
for(i=1;i<=N;i++)
{
if(p.isVisited==1)printf(" %d",i);
}
printf("\n");
}
void refresh()
{
int i,j;
for(i=0;i<SIZE;i++)
{
for(j=0;j<SIZE;j++)
{
p.paths[j]=1;
}
p.isVisited=1;
}
}
void main()
{
int N,i,e,j;
p=(vertex*)malloc(SIZE*sizeof(vertex));
while(scanf("%d",&N)==1 && N)
{
refresh();
while(scanf("%d",&i) ==1 && i)
{
while(scanf("%d",&e)==1 && e)
{
j=0;
while(p.paths[j]>=0)j++;
p.paths[j]=e;
}
}
scanf("%d",&i);
while(i>0)
{
scanf("%d",&j);
reachable=0;
doDFS(N,j);
for(e=0;e<=N;e++)
{
p[e].isVisited=1;
}
}
}
free(p);
}
280  Vertex  Floyd Warshall WA  HELP!
Hi guys, first I'd like to tell u that this one is just the second graph problem I try, so, there may exist some logic trouble. My algorithm works like this:
Define infinity to be 20000 if there is no path between two vertices.
Collect input infos and update the paths.
Use FloydWarshall to determine the shortest path between all vertices
For the desired vertex check if there's such a short path.
Output those who doesn't have it.
I've already checked and accepted all suggested inputs, but it keeps on WA. I'd be very thankful if anyone could help me.
Define infinity to be 20000 if there is no path between two vertices.
Collect input infos and update the paths.
Use FloydWarshall to determine the shortest path between all vertices
For the desired vertex check if there's such a short path.
Output those who doesn't have it.
I've already checked and accepted all suggested inputs, but it keeps on WA. I'd be very thankful if anyone could help me.
Code: Select all
#include <stdio.h>
int n, src, dst, numb, v;
int main() {
while(scanf("%d", &n) == 1 && n){
long graph [201][201];
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
graph[i][j] = 20000;
while (scanf("%d", &src) == 1 && src) {
while (scanf("%d", &dst) == 1 && dst) {
graph[src][dst] = 1;
src = dst;
}
}
/* Floyd Warshall */
for (int k = 1; k <= n; k++) {
for (int t = 1; t <= n; t++) {
for (int j = 1; j <= n; j++) {
if (graph[t][k] + graph[k][j] < graph[t][j]) {
graph[t][j] = graph[t][k] + graph[k][j];
}
}
}
}
scanf("%d", &numb);
while(numb && scanf("%d", &v) == 1){
int contador = 0;
for (int j = 1; j <= n; j++)
if (graph[v][j] >= 20000) contador++;
printf("%d", contador);
for (int j = 1; j <= n; j++)
if (graph[v][j] >= 20000) printf(" %d", j);
printf("\n");
}
}
}
UFCG Brazil  Computer Science graduate student
http://acm.uva.es/problemset/usersnew.php?user=54806 ... and going up!
http://acm.uva.es/problemset/usersnew.php?user=54806 ... and going up!