如何实现无限多维数组？ [英] How to implement infinite multidimensional array?

问题描述

我想用下面的code，我想用它来输入未知大小。比如有一个数组 INT CAC [1000] [1000] 。我可以使用矢量＆lt;矢量＆lt;＆INT GT; ＆GT;阵列; ，那我怎么才能用初始化-1 ？有什么建议？

 的#include＆LT; sstream＆GT;
＃包括LT＆;＆iostream的GT;
＃包括LT＆;矢量＆GT;
＃包括LT＆; cstdlib＆GT;
＃包括LT＆;＆memory.h GT;使用命名空间std;INT CAC [1000] [1000];
串水库[1000] [1000];
矢量＆lt;字符串＆GT;话;
INT米;INT去（INT A，INT B）{
 如果（CAC [A] [B]≥= 0）返回CAC [A] [B];
 如果（A == b）退回0; INT csum = -1;
 的for（int i为一个; I＆LT; B; ++ I）{
  csum + =词语[I] .size（）+ 1;
 }
 如果（csum＆下; = M ||一个== B-1）{
  字符串九月=;
    的for（int i为一个; I＆LT; B; ++ I）{
        RES [A] [B] .append（SEP）;
        水库[A] [B] .append（字由[i]）;
        九月=;
    }
  返回CAC [A] [B] =（M-csum）*（M-csum）;
 } INT RET = 10亿;
 INT best_sp = -1;
 对于（INT SP = A + 1; SP＆LT; B; ++ SP）{
 INT CUR =去（A，SP）+去（SP，B）;
 如果（CUR＆LT; = RET）{
    RET = CUR;
    best_sp = SP;
 }
 }
 水库[A] [B] =水库[一个] [best_sp] +\\ N+解析度[best_sp] [B];
 返回CAC [A] [B] = RET;
 }
INT主（INT ARGC，字符** argv的）{
memset的（CAC，-1，的sizeof（CAC））;
M =的atoi（ARGV [1]）;
串词;
而（CIN＆GT;＆GT;字）words.push_back（字）;
去（0，words.size（））;
COUT＆LT;＆LT;水库[0] [words.size（）]下;＆下; ENDL;
}
 

解决方案

从使用STL的std ::矢量比下面的解决方案，它指出在这篇文章的评论更为直接。我发现这个网站有效地解释了话题：的http：// www.learncpp.com/cpp-programming/16-2-stl-containers-overview/

无限大小的数组实际上不可能。但是，您可以使用动态分配基本上达到这样的效果。下面是一些示例code：

  INT计数器= 0;
为int * myArray的=新INT [1000];
 

用数据填充数组，每次添加一个值时间计数器递增。当计数器达到1000，请执行以下操作：

 为int * largerArray = INT新[2000];
的for（int i = 0; I＆LT; 1000;我++）
{
    largerArray [I] = myArray的[I]
}
删除[] myArray的;
myArray的= largerArray;
 

通过这种方法，您可以创建一个无限大小的数组最接近可能的，我不相信性能会与复制件问题

I want to use the code below and I want to use it for "unknown size of input". For example there is an array int cac[1000][1000]. I can use vector<vector<int> > array;, then how can i initialize it with -1 ? Any suggestions?

#include <sstream>
#include <iostream>
#include <vector>
#include <cstdlib>
#include <memory.h>

using namespace std;

int cac[1000][1000];
string res[1000][1000];
vector<string> words;
int M;

int go(int a, int b){
 if(cac[a][b]>= 0) return cac[a][b];
 if(a == b) return 0;

 int csum = -1;
 for(int i=a; i<b; ++i){
  csum += words[i].size() + 1;
 }
 if(csum <= M || a == b-1){
  string sep = "";
    for(int i=a; i<b; ++i){
        res[a][b].append(sep);
        res[a][b].append(words[i]);
        sep = " ";
    }
  return cac[a][b] = (M-csum)*(M-csum);
 }

 int ret = 1000000000;
 int best_sp = -1;
 for(int sp=a+1; sp<b; ++sp){
 int cur = go(a, sp) + go(sp,b);
 if(cur <= ret){
    ret = cur;
    best_sp = sp;
 }
 }
 res[a][b] = res[a][best_sp] + "\n" + res[best_sp][b];
 return cac[a][b] = ret;
 }


int main(int argc, char ** argv){
memset(cac, -1, sizeof(cac));
M = atoi(argv[1]);
string word;
while(cin >> word) words.push_back(word);
go(0, words.size());
cout << res[0][words.size()] << endl;
}

解决方案

Using std::vector from the STL is much more straightforward than the following solution, which was pointed out in the comments for this post. I find that this site explains that topic effectively: http://www.learncpp.com/cpp-programming/16-2-stl-containers-overview/

An array of infinite size is not actually possible. However, you can achieve basically that effect using dynamic allocation. Here's some sample code:

int counter = 0;
int* myArray = new int[1000];

Fill the array with data, incrementing counter each time you add a value. When counter reaches 1000, do the following:

int* largerArray = new int[2000];
for( int i = 0; i < 1000; i++ )
{
    largerArray[i] = myArray[i];
}
delete[] myArray;
myArray = largerArray;

With this method, you create the closest thing possible to an infinitely sized array, and I don't believe performance will be an issue with the copy piece

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