10973  Triangle Counting
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Hi,
I've acc but in 7,8 sec. My prog is in general O(VE) but with some optimalization. I have one question is there an O(V^2) algo or O(VE) with super optimalization?
Probably i should use this "You are assured that there is no edge in this graph that is the member of more than 2 triangles" in very smart way, but at this time i've no idea, any hints?
thx
I've acc but in 7,8 sec. My prog is in general O(VE) but with some optimalization. I have one question is there an O(V^2) algo or O(VE) with super optimalization?
Probably i should use this "You are assured that there is no edge in this graph that is the member of more than 2 triangles" in very smart way, but at this time i've no idea, any hints?
thx
Monsoon:
The first algorithm that comes to our mind is what follows:
Let M be the adjacency matrix of the graph, n the number of nodes...
int c2 = 0;
for (int i = 1; i <= n; i++)
for (int j = i + 1; j <= n; j++)
if (M[j] == 1)
{
int c1 = 0;
for (int k = j + 1; k <= n; k++)
if (M[j][k] == 1 && M[k] == 1)
{
c1++;
if (c1 == 2)
break;
}
c2 += c1;
}
printf("%d\n", c2);
The line marked in red is what you can use to make your algorithm faster..., that's why it's so important...
Now try to make the same thing in another way..., this will get TLE of course
Hope this helps,
Yandry.
The first algorithm that comes to our mind is what follows:
Let M be the adjacency matrix of the graph, n the number of nodes...
int c2 = 0;
for (int i = 1; i <= n; i++)
for (int j = i + 1; j <= n; j++)
if (M[j] == 1)
{
int c1 = 0;
for (int k = j + 1; k <= n; k++)
if (M[j][k] == 1 && M[k] == 1)
{
c1++;
if (c1 == 2)
break;
}
c2 += c1;
}
printf("%d\n", c2);
The line marked in red is what you can use to make your algorithm faster..., that's why it's so important...
Now try to make the same thing in another way..., this will get TLE of course
Hope this helps,
Yandry.
I solved it in O( VE ) but added a line which stops counting triangles for some edge after I have found 2 triangles.
It gets TLE
This is the important part of my code:
How can I make it faster?
I know it is too slow for bipartite graphs.
It gets TLE
This is the important part of my code:
Code: Select all
int cnt = 0;
for( int i = 0; i < ( int )E.size(); ++i ) {
int t_cnt = 0;
for( int j = 0; t_cnt < 2 && j < v; ++j )
t_cnt += ( path[E[i].first][j] && path[j][E[i].second] );
cnt += t_cnt;
}
I know it is too slow for bipartite graphs.
A sta da radim
i made also BFS and give all nodes v length D[v] from any start node(watch out for not connected graphs). And we can see that in triangle is at least one edge between nodes e1e2 that D[e1]==D[e2]. So we can check in your first loop only that type of edges. And we should watch out about triangle(v1,v2,v3) that D[v1]==D[v2]==D[v3].
But it gave me only time about 7 sec,
if anybody knows better algo please give me a hint?
To neno_uci:
thx for your post, but i have this condition in my solution
But it gave me only time about 7 sec,
if anybody knows better algo please give me a hint?
To neno_uci:
thx for your post, but i have this condition in my solution
This is algorithm of my solution:
It gave me about 1.5 sec runtime. (asm and faster I/O cut that down to 0.180s.)
Code: Select all
for x=1..n, adj[x] is the set { y: y>x and (x,y) is in E }.
w[1..n] is an array of integers, initially filled with zeroes
count = 0
for x = 1 to n:
for each y in adj[x]: w[y] = 1
for each y in adj[x]:
for each z in adj[y]:
count += w[z]
for each y in adj[x]: w[z] = 0
return count
10973  Triangle Counting
can anyone tell me why am i getting RE? I cant figure it out
Code: Select all
#include <cstdio>
#include <vector>
#define MAX 3002
using namespace std;
int f1[MAX], f2[MAX];
int e[MAX][2];
vector<int> a[MAX];
int main() {
int i, j, s, t, n, m, test;
long long count;
scanf("%d", &test);
while(test) {
scanf("%d %d", &n, &m);
for(i = 1; i <= n; ++i) a[i].clear();
for(i = 0; i < m; ++i) {
scanf("%d %d", &e[i][0], &e[i][1]);
a[e[i][0]].push_back(e[i][1]);
a[e[i][1]].push_back(e[i][0]);
}
memset(f1, 0, sizeof(f1));
memset(f2, 0, sizeof(f2));
count = 0;
for(i = 0; i < m; ++i) {
s = e[i][0];
t = e[i][1];
for(j = 0; j < a[s].size(); ++j) f1[a[s][j]] = i + 1;
for(j = 0; j < a[t].size(); ++j) {
f2[a[t][j]] = i + 1;
if(f1[a[t][j]] == i + 1 && a[t][j] != s && a[t][j] != t) ++count;
}
}
printf("%lld\n", count / 3);
}
return 0;
}
RTE is...
RTE is...
1. Small array size
2. Devided zero(ex : 0/0, 1/0, 0/1)
I think you are case 2.
Check!
1. Small array size
2. Devided zero(ex : 0/0, 1/0, 0/1)
I think you are case 2.
Check!

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 Location: Sanok, Poland
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As I said  you use too much memory. When std::vector fails to add a new element due to memory limit, it throws bad_alloc exception which results in RTE.
There might be some other reasons why your code fails, but this is the most important one.
There might be some other reasons why your code fails, but this is the most important one.
For millions of years, mankind lived just like the animals. Then something happened which unleashed the power of our imagination. We learned to talk and we learned to listen...

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Re: 10973  Triangle Counting
i'm getting WA!!! need some critical test cases........
Code: Select all
#include<stdio.h>
#include<math.h>
#include<ctype.h>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include<sstream>
#include<iostream>
using namespace std;
#define VI vector<int>
#define VS vector<string>
#define SZ size()
#define PB push_back
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,a,b) for(i=(a);i<(b);i++)
#define FORr(i,a,b) for(i=(a);i>=(b);i)
#define REPr(i,n) FORr(i,n1,0)
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
#define PI 2*acos(0.0)
#define EPS 1e8
bool G[3005][3005],vi[3005];
int n,tri;
void dfs(int u,int pre)
{
int v;
REP(v,n)
{
if(u==v)
continue;
if(G[u][v])
{
if(!vi[v])
{
vi[v]=true;
dfs(v,u);
}
else if(v!=pre && G[v][pre])
tri++;
}
}
}
int main()
{
int t,i,x,y,m;
scanf("%d",&t);
while(t)
{
scanf("%d %d",&n,&m);
memset(G,0,sizeof(G));
memset(vi,0,sizeof(vi));
REP(i,m)
{
scanf("%d %d",&x,&y);
x;y;
G[x][y]=G[y][x]=true;
}
tri=0;
REP(i,n)
{
if(!vi[i])
{
vi[i]=true;
dfs(i,1);
}
}
printf("%d\n",tri);
}
return 0;
}
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