Question

我的服务在4个不同的地方停运。我将每个位置中断建模为Boost ICL interval_set。我想知道什么时候至少N个地点有活动中断。

因此，在this answer之后，我实现了一个组合算法，因此我可以通过interval_set十字路口在elemenets之间创建组合。

当这个过程结束时，我应该有一定数量的interval_set，每个都同时定义N个位置的中断，最后一步将加入它们以获得所需的完整图片。

问题是我正在调试代码，当打印每个交集的时间到来时，输出文本会变得疯狂（即使我使用gdb逐步调试），我无法看到它们，导致大量CPU使用。

我想我会以某种方式发送输出比我应该更多的内存，但我无法看到问题所在。

这是一个SSCCE：

#include <boost/icl/interval_set.hpp>
#include <algorithm>
#include <iostream>
#include <vector>


int main() {
    // Initializing data for test
    std::vector<boost::icl::interval_set<unsigned int> > outagesPerLocation;
    for(unsigned int j=0; j<4; j++){
        boost::icl::interval_set<unsigned int> outages;
        for(unsigned int i=0; i<5; i++){
            outages += boost::icl::discrete_interval<unsigned int>::closed(
                (i*10), ((i*10) + 5 - j));
        }
        std::cout << "[Location " << (j+1) << "] " << outages << std::endl;
        outagesPerLocation.push_back(outages);
    }

    // So now we have a vector of interval_sets, one per location. We will combine
    // them so we get an interval_set defined for those periods where at least
    // 2 locations have an outage (N)
    unsigned int simultaneusOutagesRequired = 2;  // (N)

    // Create a bool vector in order to filter permutations, and only get
    // the sorted permutations (which equals the combinations)
    std::vector<bool> auxVector(outagesPerLocation.size());
    std::fill(auxVector.begin() + simultaneusOutagesRequired, auxVector.end(), true);

    // Create a vector where combinations will be stored
    std::vector<boost::icl::interval_set<unsigned int> > combinations;

    // Get all the combinations of N elements
    unsigned int numCombinations = 0;
    do{
        bool firstElementSet = false;
        for(unsigned int i=0; i<auxVector.size(); i++){
            if(!auxVector[i]){
                if(!firstElementSet){
                    // First location, insert to combinations vector
                    combinations.push_back(outagesPerLocation[i]);
                    firstElementSet = true;
                }
                else{
                    // Intersect with the other locations
                    combinations[numCombinations] -= outagesPerLocation[i];
                }
            }
        }
        numCombinations++;
        std::cout << "[-INTERSEC-] " << combinations[numCombinations] << std::endl;  // The problem appears here
    }
    while(std::next_permutation(auxVector.begin(), auxVector.end()));

    // Get the union of the intersections and see the results
    boost::icl::interval_set<unsigned int> finalOutages;
    for(std::vector<boost::icl::interval_set<unsigned int> >::iterator
        it = combinations.begin(); it != combinations.end(); it++){
        finalOutages += *it;
    }

    std::cout << finalOutages << std::endl;
    return 0;
}

任何帮助？

Answer 1

作为I surmised，有一个＆＃34;高级＆＃34;接近这里。

提升ICL容器不仅仅是一对间隔开始/结束点和＃34;的美化容器。它们旨在以一般优化的方式实现仅仅组合，搜索的业务。

所以你不必。

如果你让图书馆做他们应该做的事情：

using TimePoint = unsigned;
using DownTimes = boost::icl::interval_set<TimePoint>;
using Interval  = DownTimes::interval_type;
using Records   = std::vector<DownTimes>;

使用功能域typedef邀请更高级别的方法。现在，让我们问一下假设的商业问题＆＃34;：

我们对每个位置停机记录的实际想做些什么？

好吧，我们基本上想要

计算所有可辨别的时段和
过滤那些标记至少为2的文件
最后，我们要展示＆＃34;合并＆＃34;留下的时间段。

好的，工程师：实施它！

嗯。清点。它能有多难？

elegant优雅解决方案的关键是选择正确的数据结构

using Tally     = unsigned; // or: bit mask representing affected locations?
using DownMap   = boost::icl::interval_map<TimePoint, Tally>;

现在它只是批量插入：

// We will do a tally of affected locations per time slot
DownMap tallied;
for (auto& location : records)
    for (auto& incident : location)
        tallied.add({incident, 1u});

好的，让我们过滤一下。我们只需要适用于我们的DownMap的谓词，对吧

// define threshold where at least 2 locations have an outage
auto exceeds_threshold = [](DownMap::value_type const& slot) {
    return slot.second >= 2;
};

合并时段！

实际上。我们只是创建另一个DownTimes集。只是，这次不是每个地点。

数据结构的选择再次获胜：

// just printing the union of any criticals:
DownTimes merged;
for (auto&& slot : tallied | filtered(exceeds_threshold) | map_keys)
    merged.insert(slot);

报告！

std::cout << "Criticals: " << merged << "\n";

请注意，我们无处接近操纵数组索引，重叠或非重叠间隔，闭合或开放边界。或者，[eeeeek！]收集元素的强力排列。

我们刚刚阐述了目标，让图书馆开展工作。

完整演示

<强> Live On Coliru

#include <boost/icl/interval_set.hpp>
#include <boost/icl/interval_map.hpp>
#include <boost/range.hpp>
#include <boost/range/algorithm.hpp>
#include <boost/range/adaptors.hpp>
#include <boost/range/numeric.hpp>
#include <boost/range/irange.hpp>
#include <algorithm>
#include <iostream>
#include <vector>

using TimePoint = unsigned;
using DownTimes = boost::icl::interval_set<TimePoint>;
using Interval  = DownTimes::interval_type;
using Records   = std::vector<DownTimes>;

using Tally     = unsigned; // or: bit mask representing affected locations?
using DownMap   = boost::icl::interval_map<TimePoint, Tally>;

// Just for fun, removed the explicit loops from the generation too. Obviously,
// this is bit gratuitous :)
static DownTimes generate_downtime(int j) {
    return boost::accumulate(
            boost::irange(0, 5),
            DownTimes{},
            [j](DownTimes accum, int i) { return accum + Interval::closed((i*10), ((i*10) + 5 - j)); }
        );
}

int main() {
    // Initializing data for test
    using namespace boost::adaptors;
    auto const records = boost::copy_range<Records>(boost::irange(0,4) | transformed(generate_downtime));

    for (auto location : records | indexed()) {
        std::cout << "Location " << (location.index()+1) << " " << location.value() << std::endl;
    }

    // We will do a tally of affected locations per time slot
    DownMap tallied;
    for (auto& location : records)
        for (auto& incident : location)
            tallied.add({incident, 1u});

    // We will combine them so we get an interval_set defined for those periods
    // where at least 2 locations have an outage
    auto exceeds_threshold = [](DownMap::value_type const& slot) {
        return slot.second >= 2;
    };

    // just printing the union of any criticals:
    DownTimes merged;
    for (auto&& slot : tallied | filtered(exceeds_threshold) | map_keys)
        merged.insert(slot);

    std::cout << "Criticals: " << merged << "\n";
}

打印

Location 1 {[0,5][10,15][20,25][30,35][40,45]}
Location 2 {[0,4][10,14][20,24][30,34][40,44]}
Location 3 {[0,3][10,13][20,23][30,33][40,43]}
Location 4 {[0,2][10,12][20,22][30,32][40,42]}
Criticals: {[0,4][10,14][20,24][30,34][40,44]}

Answer 2

在置换循环结束时，你写道：

numCombinations++;
std::cout << "[-INTERSEC-] " << combinations[numCombinations] << std::endl;  // The problem appears here

我的调试器告诉我，在第一次迭代numCombinations 在增量之前是 0。但是增加它会使它超出combinations容器的范围（因为它只是一个元素，所以索引为0）。

您的意思是在使用后增加它吗？是否有任何特殊原因不使用

std::cout << "[-INTERSEC-] " << combinations.back() << "\n";

或者，对于c ++ 03

std::cout << "[-INTERSEC-] " << combinations[combinations.size()-1] << "\n";

甚至只是：

std::cout << "[-INTERSEC-] " << combinations.at(numCombinations) << "\n";

会抛出std::out_of_range？

另一方面，我认为Boost ICL有 更有效的方法来获得你想要的答案。让我考虑一下这一点。如果我看到它会发布另一个答案。

更新：使用Boost ICL发布 other answer show casing highlevel编码

N Boost interval_set的组合

2 个答案:

所以你不必。

完整演示