11307 - Alternative Arborescence

[mmonish]

can anyone give me some hint on how to solve this problem..

[sclo]

DP:

let f(i,j) = minimum sum of subtree rooted at i, where i has color j

Further, the maximum color we need can be more than 2

[rushel]

how stupid i am. i tried dp with 3 colors. thank sclo now i got ACCEPTED using 20 colors.

but how could i proof that how many colors are needed.

[mmonish]

I tried DP but i am getting WA. Anyone please give me some critical test case..

[baodog]

I believe number of colors can be bounded by C*log n or so.
In this case exactly 6 colors suffice.

[mmonish]

Can anyone tell me what i did wrong here..

Code: Select all

Remove After AC

[baodog]

Looks like you assume root is 0.
The root is in general not 0.

[mmonish]

>>baodog
Thx for ur quick reply. Now i get AC.

[goodwill]

You may find this O(n) algorithm interesting (no need to know how many color there are).
http://citeseer.ist.psu.edu/494676.html

[Darko]

I found the way to build the minimum size trees that need k colors and the linear algorithm to calculate the chromatic sum of a tree in the paper referenced there:
'Introduction to chromatic sums' - Kubicka, Schwenk (1989)
Expanded version is the first section of Kubicka's PhD thesis.

I thought it was fun, just because it is not that obvious how many colors you need and some obvious greedy approaches don't work (like assigning 1's to a max independent set).

A tree needs more than 10000 nodes to use the 9th color. The sequence of minimum number of nodes that force kth color is in OEIS - I didn't find the chromatic sum mentioned there, but that is the sequence.

baodog is right, test cases go only up to 6 colors (I was doing them by hand - too lazy, I guess, but if you solve for 6 I think you solved it).

[Robert Gerbicz]

You may find this O(n) algorithm interesting (no need to know how many color there are).
http://citeseer.ist.psu.edu/494676.html

Today I've implemented that algorithm. Giving me the first position on this problem's ranklist by only 0.030 sec. of running time.

[sclo]

Actually, on the contest, I saw many people got WA the first try, so I guessed that greedy doesn't really work, and tried an solution for 10 colors, since it seems reasonable that the number of color is around O(log n)

[Donotalo]

I tried with 10 colors but got WA. Here is my code:

Code: Select all

#include <iostream>
#include <string>
#include <sstream>
using namespace std;

#define MAX 10004
#define NCOLOR 10
unsigned long INF = (1 << 29);

int degree[MAX], graph[MAX][MAX], cost[MAX][NCOLOR];

int main()
{
   register int m, p;
   int n, i, j, k, color;
   string line, first;

   while(cin >> n) {
      if (n == 0) break;

      cin.ignore();
      memset(degree, 0, sizeof(degree));
      memset(graph, 0, sizeof(graph));

      for (i = 0; i < n; i++) {
         getline(cin, line);
         stringstream sstr;
         sstr << line;
         sstr >> first;
         first.erase(first.size()-1, 1);
         stringstream sstr2;
         sstr2 << first;
         sstr2 >> j;

         while(sstr >> k) {
            if (k > j)
               graph[j][degree[j]++] = k;
            else
               graph[k][degree[k]++] = j;
         }
      }

/*      for (i = 0; i < n; i++) {
         cout << i << ": ";
         for (j = 0; j < degree[i]; j++)
            cout << graph[i][j] << ' ';
         cout << endl;
      }
      cout << endl;*/

      for (i = n-1; i >= 0; i--) {
         if (degree[i] == 0) {
            for (j = 0; j < NCOLOR; j++)
               cost[i][j] = j + 1;
         }
         else {
            for (color = 0; color < NCOLOR; color++) {
               k = 0;
               for (j = 0; j < degree[i]; j++) {
                  m = INF;
                  for (p = 0; p < NCOLOR; p++) {
                     if (p == color) continue;
                     if (cost[graph[i][j]][p] < m)
                        m = cost[graph[i][j]][p];
                  }
                  k += m;
               }
               cost[i][color] = k + color + 1;
            }
         }
      }

      m = cost[0][0];
      for (i = 1; i < degree[0]; i++)
         if (cost[0][i] < m)
            m = cost[0][i];
      cout << m << endl;
   }

   return 0;
}

Can anyone help me to find my mistake?

[Sedefcho]

Hello all,

I have implemented the algorithm described in the OCCP article
by Leo G. Kroon, Arunabha Sen, Haiyong Deng and Asim Roy.

Check this link (it's a PostScript file):
http://www.public.asu.edu/~halla/papers/OCCPWG96.ps

I have tested it with several test cases (simple ones)
and the answers seem OK.
I got RE from the judge though.

Could someone give any hint about special cases,
boundary conditions etc. I am stuck as the new system
provides no feedback at all about the error (not even
a simple email).

Thanks in advance.

Sedefcho

[lonelyone]

Why it got Runtime Error? Could someone explain this for me? Thanks a lot.

Code: Select all

for(i=0; i<n; ++i)
        {
            gets(line);
            p = line;
            sscanf(p, "%d%n", &from, &len);
            p += len;
            while(*p)
            {
                if(sscanf(p, "%d%n", &to, &len) == 1)
                {
                     v[from].push_back(to);
                }
                p += len;
            }
        }