递归算法无法在指定时间内完成测试

本文介绍了递归算法无法在指定时间内完成测试的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在做的测试需要二进制层析成像算法.提供一组38个测试值来测试正确性，但是完成所有测试也有1个CPU秒的时间限制.问题如下:

I was doing a test that required an algorithm for Binary Tomography. A set of 38 test values are supplied that test correctness, but there is also a time limit of 1 CPU sec to complete all the tests. The problem is as follows:

如果存在一个m×n矩阵A，且每个元素为0或1，则输出是"，这样

Output "Yes" if there exists an m-by-n matrix A, with each element either being 0 or 1, such that

否则输出否".

为每个测试提供2个阵列:

For each test, 2 arrays are provided:

r (矩阵中每一行的总和)

c (矩阵中每一列的总和)

r (the sum of each row in the matrix)

c (the sum of each column in the matrix)

在等式中:

• m 是 r 数组的长度，其中 1< = m
• n 是 c 数组的长度，其中 n< = 1000
• ri 是 r 的元素，其中 0< = ri< = n
• cj 是 c 的元素，其中 0< = cj< = m

• m is the length of the r array, where 1 <= m
• n is the length of the c array, where n <= 1000
• ri is an element of r, where 0 <= ri <= n
• cj is an element of c, where 0 <= cj <= m

是"示例

m = 3;n = 4；r = [2，3，2];c = [1、1、3、2];

m = 3; n = 4; r = [2, 3, 2]; c = [1, 1, 3, 2];

否"示例

m = 3;n = 3；r = [0，0，3];c = [0，0，3];

m = 3; n = 3; r = [0, 0, 3]; c = [0, 0, 3];

我有一个解决方案似乎可以给出正确的答案，但是它只能在超过1秒的CPU时间之前管理12/38个测试.

I have a solution that appears to give correct answers, however it only manages 12 / 38 tests before the 1 second of CPU time is exceeded.

我最初是在ES5中编写代码，然后返回并转换为ES3，以尝试从中获得更多性能.(最初将9个测试作为ES5管理).似乎没有什么可以改进当前算法的性能了(除非我弄错了).这使我相信我的算法是错误的，并且必须有一个更快的算法来执行此操作.我做了很多阅读，试图找到一个，结果很头疼:)

I originally wrote the code in ES5 and then went back and converted to to ES3 to try and get more performance out of it. (originally managed 9 tests as ES5). There doesn't seem a great deal left that I can do to the current algorithm to improve the performance (unless I am mistaken). This leads me to believe that my algorithm is at fault an that there must be a faster algorithm for doing this. I did a ton of reading trying to find one and ended up with a headache :)

因此，我正在求助于社区，看看是否有人可以提出比我目前正在使用的算法更快的算法.

So I'm turning to the community to see if anyone can suggest a faster algorithm than I am currently using.

'use strict'; const ZEROS = (function (seed) { let string = seed; for (let i = 0; i < 19; i += 1) { string += seed; } return string; }('00000000000000000000000000000000000000000000000000')); const ZEROSLEN = ZEROS.length; const permutate = function (n, ri) { const result = []; const memoize = {}; let count = 0; do { const bin = count.toString(2); if (ZEROSLEN + bin.length > ZEROSLEN + n) { break; } if (!memoize[bin] && (bin.split('1').length - 1) === ri) { const string = (ZEROS + bin).slice(-n); const sLen = string.length; const perm = new Array(sLen); for (let i = sLen - 1; i >= 0; i -= 1) { perm[i] = +string[i]; } memoize[bin] = result.push(perm); } count += 1; } while (count); return result; }; const getMatrixSum = function (n, matrix) { const mLength = matrix.length; const rows = new Array(mLength); const a = new Array(n); const last = mLength - 1; for (let x = n - 1; x >= 0; x -= 1) { for (let y = last; y >= 0; y -= 1) { rows[y] = matrix[y][x]; } let sum = 0; for (let i = rows.length - 1; i >= 0; i -= 1) { sum += rows[i]; } a[x] = sum; } return a; }; const isEqual = function (a, b) { const length = a.length; if (length !== b.length) { return false; } for (let i = length - 1; i >= 0; i -= 1) { if (a[i] !== b[i]) { return false; } } return true; }; const addRow = function (i, prev, r, c, result) { if (result) { return result; } const n = c.length; const ri = r[i]; if (ri < 0 || ri > n) { throw new RangeError('ri out of range'); } const p = permutate(n, ri); const m = r.length; const rsLast = m - 1; const nextI = i + 1; for (let x = p.length - 1; x >= 0; x -= 1) { const permutation = p[x]; const next = prev.slice(); next.push(permutation); const sums = getMatrixSum(n, next); if (i < rsLast) { let memo = 0; for (let j = sums.length - 1; j >= 0; j -= 1) { if (sums[j] > c[j]) { memo += 1; } } if (!memo && addRow(nextI, next, r, c, result)) { return true; } } else if (isEqual(sums, c)) { return true; } } return false; }; const isSolvable = function (r, c) { const m = r.length; const n = c.length; if (m < 1 || n > 1000) { throw new Error('Bad data'); } for (let j = n; j >= 0; j -= 1) { const cj = c[j]; if (cj < 0 || cj > m) { throw new RangeError('cj out of range'); } } return addRow(0, [], r, c, false) ? 'Yes' : 'No'; }; console.log(isSolvable([2, 3, 2], [1, 1, 3, 2])); console.log(isSolvable([0, 0, 3], [0, 0, 3]));

可能值得注意的是，这些测试是在SpiderMonkey版本JavaScript-C24.2.0上运行的

It may be worth noting that the tests are being run on SpiderMonkey version JavaScript-C24.2.0

参考:

en.wikipedia/wiki/Discrete_tomography

open.kattis/problems/tomography

推荐答案

我还没有准备好进行测试，但是事件发生后我发现了一种效率更高的算法.

I didn't have this ready for my test, but I found a far more efficient algorithm after the event.

'use strict'; const sortNumber = function (a, b) { return b - a; }; const isSolvable = function (r, c) { const m = r.length; const n = c.length; if (m < 1 || n > 1000) { throw new Error('Bad data'); } for (let j = n; j >= 0; j -= 1) { const cj = c[j]; if (cj < 0 || cj > m) { throw new RangeError('cj out of range'); } } while (r.length) { c.sort(sortNumber); const ri = r.pop(); if (ri < 0 || ri > n) { throw new RangeError('ri out of range'); } if (ri) { if (!c[ri - 1]) { return 'No'; } for (let j = ri - 1; j >= 0; j -= 1) { c[j] -= 1; } } } for (let j = n - 1; j >= 0; j -= 1) { if (c[j]) { return 'No'; } } return 'Yes'; }; console.log(isSolvable([2, 3, 2], [1, 1, 3, 2])); console.log(isSolvable([0, 0, 3], [0, 0, 3]));