Question

假设我们得到一组数字，如20 40 20 60 80 60 它可以分为2个回文序列：20 40 20和60 80 60。它也可以分为6个回文序列，每个序列包含一个数字。

我们如何在c ++中找到一组给定数字中最小数量的回文序列？

PS-这不是我的功课。真正的问题。

Answer 1

通过查看每个O（n ³）子序列并检查它是否是回文，开始直接的方法。一旦我们知道哪些子序列是回文，我们就可以在O（n ²）时间内进行动态编程，以找到覆盖整个序列的最小连续子序列数。

对于输入20 40 20 60 80 60，下面的C ++实现打印[20 40 20] [60 80 60]。

#include <cstdio>
#include <vector>
using namespace std;

int main() {
  // Read the data from standard input.
  vector<int> data;
  int x;
  while (scanf("%d", &x) != EOF) {
    data.push_back(x);
  }
  int n = data.size();

  // Look at every subsequence and determine if it's a palindrome.
  vector<vector<bool> > is_palindrome(n);
  for (int i = 0; i < n; ++i) {
    is_palindrome[i] = vector<bool>(n);
    for (int j = i; j < n; ++j) {
      bool okay = true;
      for (int left = i, right = j; left < right; ++left, --right) {
        if (data[left] != data[right]) { 
          okay = false;
          break; 
        }
      }
      is_palindrome[i][j] = okay;
    }
  }

  // Dynamic programming to find the minimal number of subsequences.
  vector<pair<int,int> > best(n);
  for (int i = 0; i < n; ++i) {
    // Check for the easy case consisting of one subsequence.
    if (is_palindrome[0][i]) {
      best[i] = make_pair(1, -1);
      continue;
    }
    // Otherwise, make an initial guess from the last computed value.
    best[i] = make_pair(best[i-1].first + 1, i-1);
    // Look at earlier values to see if we can improve our guess.
    for (int j = i-2; j >= 0; --j) {
      if (is_palindrome[j+1][i]) {
        if (best[j].first + 1 < best[i].first) {
          best[i].first = best[j].first + 1;
          best[i].second = j;
        }
      }
    }
  }

  printf("Minimal partition: %d sequences\n", best[n-1].first);
  vector<int> indices;
  int pos = n-1;
  while (pos >= 0) {
    indices.push_back(pos);
    pos = best[pos].second;
  }
  pos = 0;
  while (!indices.empty()) {
    printf("[%d", data[pos]);
    for (int i = pos+1; i <= indices.back(); ++i) {
      printf(" %d", data[i]);
    }
    printf("] ");
    pos = indices.back()+1;
    indices.pop_back();
  }
  printf("\n");

  return 0;
}

从给定的一组数字中找出回文序列的数量

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