greed.cc.txt

来自「Ulm大学2003-2004年竞赛题」· 文本 代码 · 共 62 行

// Problem   Huffman's Greed
// Algorithm Dynamic Programming
// Runtime   O(n^3)
// Author    Walter Guttmann
// Date      2003.12.07

#include <cassert>
#include <fstream>
#include <iostream>

using namespace std;

ifstream in("greed.in");

// ps[i] == q[0]+p[1]+q[1]+...+p[i]+q[i] (i.e., partial sums of the probabilities).
int ps[256];

// i<=j ==> dp[i][j] == optimum cost of arranging the labels K[i+1],...,K[j] in a tree.
// The corresponding probabilities are q[i],p[i+1],q[i+1],...,p[j],q[j].
int dp[256][256];

int main()
{
  int n;
  while (in >> n)
  {
    if (n == 0)
      break;
    assert(1 <= n && n <= 200);
    int p[256], q[256];
    for (int i=1 ; i<=n ; i++)
      in >> p[i];
    for (int i=0 ; i<=n ; i++)
      in >> q[i];

    // Calculate the partial sums.
    ps[0] = q[0];
    for (int i=1 ; i<=n ; i++)
      ps[i] = ps[i-1] + p[i] + q[i];

    // Calculate the optimum costs in the order of increasing difference d = j-i.
    for (int i=0 ; i<=n ; i++)
      dp[i][i] = 0;
    for (int d=1 ; d<=n ; d++)
      for (int i=0, j=i+d ; j<=n ; i++, j++)
      {
        // Any of K[i+1],...,K[j] may be at the root of the tree containing these labels.
        // This is the special case K[i+1]:
        dp[i][j] = dp[i+1][j];
        // These are the other cases K[i+2],...,K[j]:
        for (int k=i+2 ; k<=j ; k++)
          dp[i][j] = min(dp[i][j], dp[i][k-1] + dp[k][j]);
        // Finally, this constant amount arises in all cases considered:
        dp[i][j] += ps[j] - ps[i] + q[i];
      }

    cout << dp[0][n] << endl;
  }
  return 0;
}