在给定行和列总和的情况下查找二进制矩阵是否存在

问题描述

如何确定是否可以构造具有给定行和列总和的二进制矩阵.

How to find out if it is possible to contruct a binary matrix with given row and column sums.

输入:

输入的第一行包含两个数字1≤m，n≤1000，即矩阵的行数和列数.下一行包含m个数字0≤ri≤n–矩阵中每一行的总和.第三行包含n个数字0≤cj≤m–矩阵中每一列的总和.

The first row of input contains two numbers 1≤m,n≤1000, the number of rows and columns of the matrix. The next row contains m numbers 0≤ri≤n – the sum of each row in the matrix. The third row contains n numbers 0≤cj≤m – the sum of each column in the matrix.

输出:

如果存在一个m×n矩阵A(每个元素为0或1)，则输出"YES".否则为"NO".

Output "YES" if there exists an m-by-n matrix A, with each element either being 0 or 1. Else "NO".

我试图阅读有关层析成像算法的文章，但由于与层析成像算法有关的所有论文都非常复杂，因此无法找到答案.

I tried reading about Tomography algorithms but could not figure out an answer as all the papers related to Tomography algorithm is very complicated.

有人可以帮我吗?.

推荐答案

我已经成功实现了使用基于网络流.我打算有一天写下这些想法，但还没有找到时间.为此，我读了在场—不在场矩阵的随机化:评论和新算法米克洛斯(Miklós)和波达尼(Podani):

I've successfully implemented randomly generating such matrices for R using a modeling based on network flow. I intend to write up those ideas one day, but haven't found the time yet. Reasearching for that, I read in Randomization of Presence–absence Matrices: Comments and New Algorithms by Miklós and Podani:

我想这应该是递归确定问题的最佳方法.

I guess that should be the best method to recursively decide your question.

用我自己的话来说:选择任意行，将其从总计列表中删除.呼叫删除的号码 k .还要从 k 列中减去一个大和.您可以获得一个较小矩阵的描述，然后递归.如果在任何时候您都没有 k 列的总和不为零，那么就不会存在这样的矩阵.否则，您可以使用相反的过程来递归地建立一个匹配矩阵:取递归调用返回的矩阵，然后再添加一个带有 k 个行，并排在您最初从中减去一列的列中.

Phrased in my own words: Choose any row, remove it from the list of totals. Call that removed number k. Also subtract one from the k columns with larges sums. You obtain a description of a smaller matrix, and recurse. If at any point you don't have k columns with non-zero sums, then no such matrix can exist. Otherwise you can recursively build a matching matrix using the reverse process: take the matrix returned by the recursive call, then add one more row with k ones, placed in the columns from whose counts you originally subtracted one.

bool satisfiable(std::vector<int> a, std::vector<int> b) {
  while (!a.empty()) {
    std::sort(b.begin(), b.end(), std::greater<int>());
    int k = a.back();
    a.pop_back();
    if (k > b.size()) return false;
    if (k == 0) continue;
    if (b[k - 1] == 0) return false;
    for (int i = 0; i < k; i++)
      b[i]--;
  }
  for (std::vector<int>::iterator i = b.begin(); i != b.end(); i++)
    if (*i != 0)
      return false;
  return true;
}

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