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### Re: 11307 - Alternative Arborescence

Posted: **Thu Jun 09, 2011 10:53 pm**

by **Shafaet_du**

7 colors gave me ac,5 color wa. But i think a tree should have no loops and chromatic number should be 2. but this problem statement says there can be loops.

Don't assume 0 as root. The node which have no in-degree is the root.

### Re: 11307 - Alternative Arborescence

Posted: **Wed Jun 15, 2011 3:40 pm**

by **Imti**

I'm confused about the statement of this problem.As I can see here in this thread,everyone solved this problem with more color than 2.But Problem says..

you will be given a tree (connected graph with no simple loops)

There will be no simple loop and will be connected and we also know a tree must have n-1 edges.More than n-1 edge implies that this graph is not a tree.Like

following graph with 3 node and 3 edge is not a tree.

1 2

1 3

2 3

So,as shafaet said,this problem should be solvable with 2 color and shouldn't have any loop.May be I haven't understood problem properly.Can anyone help me

with this?

### Re: 11307 - Alternative Arborescence

Posted: **Wed Dec 26, 2012 11:16 am**

by **illusi0nist**

Imti wrote:I'm confused about the statement of this problem.As I can see here in this thread,everyone solved this problem with more color than 2.But Problem says..

you will be given a tree (connected graph with no simple loops)

There will be no simple loop and will be connected and we also know a tree must have n-1 edges.More than n-1 edge implies that this graph is not a tree.Like

following graph with 3 node and 3 edge is not a tree.

1 2

1 3

2 3

So,as shafaet said,this problem should be solvable with 2 color and shouldn't have any loop.May be I haven't understood problem properly.Can anyone help me

with this?

Well, you are right about the fact that a tree is 2-colorable. But the question isn't asking you to find the minimum number of colors, rather its asking for the minimum cost. Try coming up with a small example where using 3 colors could get you a smaller cost than 2 colors. This will convince you about the question.

On the other hand I've been stuck on this question for some time, getting WA. I believe my algorithm is right (dp with 20 colors), and it works on all cases I've tried (but they were small). So what could be wrong?

Code: Select all

```
My bad. The dynamic structure was wrong earlier. Got AC
```

### Re:

Posted: **Mon Feb 25, 2013 8:50 am**

by **xueguoqing**

baodog wrote:I believe number of colors can be bounded by C*log n or so.

In this case exactly 6 colors suffice.

But can you tell me why??

### Re: 11307 - Alternative Arborescence

Posted: **Sat May 24, 2014 9:37 am**

by **prashantharshi**

here i used dp algorithm

still i am getting time limit exceeded

here is my code

http://ideone.com/xHJ0FI
i need help if any crucial point is there in algo

### Re: 11307 - Alternative Arborescence

Posted: **Tue May 27, 2014 9:37 am**

by **@li_kuet**

I have already solved the problem.

But i'm still wondering why it is necessary to use more than two colors to color a tree

As we know a tree is Bi-Colorable, so wouldn't it be enough to use only two color ??

Can anyone give me an example where using three or more color is optimal than using two color... cause i couldn't come up with one

### Re: 11307 - Alternative Arborescence

Posted: **Thu Jun 12, 2014 10:28 pm**

by **brianfry713**

prashantharshi, 6 colors is enough.

For input:

Code: Select all

```
11
0: 3 2 7
1:
2:
3:
4: 0 10 8 6
5:
6:
7:
8: 9
9: 1 5
10:
0
```

Using 2 colors the best you can do is 16, using 3 or more the best you can do is 15.

I got AC always using node 0 as the root, and couldn't find a test case where using another root got a better result.

### Re: 11307 - Alternative Arborescence

Posted: **Sat Jun 14, 2014 5:19 am**

by **@li_kuet**

Thanks brianfry713

Now i understand, why more than 2 colors needed