干货|迭代局部搜索算法(Iterated local search)探幽（附C++代码及注释）

////////////////////////
//TSP问题 迭代局部搜索求解代码
//基于Berlin52例子求解
//作者：infinitor
//时间：2018-04-12
////////////////////////


#include <iostream>
#include <cmath>
#include <stdlib.h>
#include <time.h>
#include <vector>
#include <windows.h>
#include <memory.h>
#include <string.h>
#include <iomanip>

#define DEBUG

using namespace std;

#define CITY_SIZE 52 //城市数量


//城市坐标
typedef struct candidate
{
    int x;
    int y;
}city, CITIES;

//优化值
int **Delta; 

//解决方案
typedef struct Solution
{
    int permutation[CITY_SIZE]; //城市排列
    int cost;                        //该排列对应的总路线长度
}SOLUTION;
// 计算邻域操作优化值 
int calc_delta(int i, int k, int *tmp, CITIES * cities);

//计算两个城市间距离
int distance_2city(city c1, city c2);

//根据产生的城市序列，计算旅游总距离
int cost_total(int * cities_permutation, CITIES * cities);

//获取随机城市排列, 用于产生初始解
void random_permutation(int * cities_permutation);

//颠倒数组中下标begin到end的元素位置, 用于two_opt邻域动作
void swap_element(int *p, int begin, int end);

//邻域动作 反转index_i <-> index_j 间的元素
void two_opt_swap(int *cities_permutation, int *new_cities_permutation, int index_i, int index_j);

//本地局部搜索，边界条件 max_no_improve
void local_search(SOLUTION & best, CITIES * cities, int max_no_improve);



//将城市序列分成4块，然后按块重新打乱顺序。
//用于扰动函数
void double_bridge_move(int *cities_permutation, int * new_cities_permutation);

//扰动
void perturbation(CITIES * cities, SOLUTION &best_solution, SOLUTION &current_solution);

//迭代搜索
void iterated_local_search(SOLUTION & best, CITIES * cities, int max_iterations, int max_no_improve);

// 更新Delta 
void Update(int i, int k,  int *tmp, CITIES * cities);

//城市排列
int permutation[CITY_SIZE];
//城市坐标数组
CITIES cities[CITY_SIZE];


//berlin52城市坐标，最优解7542好像
CITIES berlin52[CITY_SIZE] = { { 565,575 },{ 25,185 },{ 345,750 },{ 945,685 },{ 845,655 },
{ 880,660 },{ 25,230 },{ 525,1000 },{ 580,1175 },{ 650,1130 },{ 1605,620 },
{ 1220,580 },{ 1465,200 },{ 1530,5 },{ 845,680 },{ 725,370 },{ 145,665 },
{ 415,635 },{ 510,875 },{ 560,365 },{ 300,465 },{ 520,585 },{ 480,415 },
{ 835,625 },{ 975,580 },{ 1215,245 },{ 1320,315 },{ 1250,400 },{ 660,180 },
{ 410,250 },{ 420,555 },{ 575,665 },{ 1150,1160 },{ 700,580 },{ 685,595 },
{ 685,610 },{ 770,610 },{ 795,645 },{ 720,635 },{ 760,650 },{ 475,960 },
{ 95,260 },{ 875,920 },{ 700,500 },{ 555,815 },{ 830,485 },{ 1170,65 },
{ 830,610 },{ 605,625 },{ 595,360 },{ 1340,725 },{ 1740,245 } };

int main()
{
    srand(1);
    int max_iterations = 600;
    int max_no_improve = 50;
    //初始化指针数组 
    Delta = new int*[CITY_SIZE];
    for (int i = 0; i < CITY_SIZE; i ++)
        Delta[i] = new int[CITY_SIZE];

    SOLUTION best_solution;

    iterated_local_search(best_solution, berlin52, max_iterations, max_no_improve);

    cout << endl<<endl<<"搜索完成！ 最优路线总长度 = " << best_solution.cost << endl;
    cout << "最优访问城市序列如下：" << endl;
    for (int i = 0; i < CITY_SIZE;i++)
    {
        cout << setw(4) << setiosflags(ios::left) << best_solution.permutation[i];
    }

    cout << endl << endl;

    return 0;
}



//计算两个城市间距离
int distance_2city(city c1, city c2)
{
    int distance = 0;
    distance = sqrt((double)((c1.x - c2.x)*(c1.x - c2.x) + (c1.y - c2.y)*(c1.y - c2.y)));

    return distance;
}

//根据产生的城市序列，计算旅游总距离
//所谓城市序列，就是城市先后访问的顺序，比如可以先访问ABC，也可以先访问BAC等等
//访问顺序不同，那么总路线长度也是不同的
//p_perm 城市序列参数
int cost_total(int * cities_permutation, CITIES * cities)
{
    int total_distance = 0;
    int c1, c2;
    //逛一圈，看看最后的总距离是多少
    for (int i = 0; i < CITY_SIZE; i++)
    {
        c1 = cities_permutation[i];
        if (i == CITY_SIZE - 1) //最后一个城市和第一个城市计算距离
        {
            c2 = cities_permutation[0];
        }
        else
        {
            c2 = cities_permutation[i + 1];
        }
        total_distance += distance_2city(cities[c1], cities[c2]);
    }

    return total_distance;
}

//获取随机城市排列
void random_permutation(int * cities_permutation)
{
    int i, r, temp;
    for (i = 0; i < CITY_SIZE; i++)
    {
        cities_permutation[i] = i; //初始化城市排列，初始按顺序排
    }


    for (i = 0; i < CITY_SIZE; i++)
    {
        //城市排列顺序随机打乱
        r = rand() % (CITY_SIZE - i) + i;
        temp = cities_permutation[i];
        cities_permutation[i] = cities_permutation[r];
        cities_permutation[r] = temp;
    }
}




//颠倒数组中下标begin到end的元素位置
void swap_element(int *p, int begin, int end)
{
    int temp;
    while (begin < end)
    {
        temp = p[begin];
        p[begin] = p[end];
        p[end] = temp;
        begin++;
        end--;
    }
}


//邻域动作 反转index_i <-> index_j 间的元素
void two_opt_swap(int *cities_permutation, int *new_cities_permutation, int index_i, int index_j)
{
    for (int i = 0; i < CITY_SIZE; i++)
    {
        new_cities_permutation[i] = cities_permutation[i];
    }

    swap_element(new_cities_permutation, index_i, index_j);
}



int calc_delta(int i, int k,  int *tmp, CITIES * cities){
    int delta = 0;
    /*
                以下计算说明：
                对于每个方案，翻转以后没必要再次重新计算总距离
                只需要在翻转的头尾做个小小处理

                比如：
                有城市序列   1-2-3-4-5 总距离 = d12 + d23 + d34 + d45 + d51 = A
                翻转后的序列 1-4-3-2-5 总距离 = d14 + d43 + d32 + d25 + d51 = B
                由于 dij 与 dji是一样的，所以B也可以表示成 B = A - d12 - d45 + d14 + d25
                下面的优化就是基于这种原理
    */
    if (i == 0)
    {
        if (k == CITY_SIZE - 1)
        {
           delta = 0;
        }
        else
        {
            delta = 0
                - distance_2city(cities[tmp[k]], cities[tmp[k + 1]])
                + distance_2city(cities[tmp[i]], cities[tmp[k + 1]])
                - distance_2city(cities[tmp[CITY_SIZE - 1]], cities[tmp[i]])
                + distance_2city(cities[tmp[CITY_SIZE - 1]], cities[tmp[k]]);
        }

    }
    else
    {
        if (k == CITY_SIZE - 1)
        {
            delta = 0
                - distance_2city(cities[tmp[i - 1]], cities[tmp[i]])
                + distance_2city(cities[tmp[i - 1]], cities[tmp[k]])
                - distance_2city(cities[tmp[0]], cities[tmp[k]])
                + distance_2city(cities[tmp[i]], cities[tmp[0]]);
        }
        else
        {
            delta = 0
                - distance_2city(cities[tmp[i - 1]], cities[tmp[i]])
                + distance_2city(cities[tmp[i - 1]], cities[tmp[k]])
                - distance_2city(cities[tmp[k]], cities[tmp[k + 1]])
                + distance_2city(cities[tmp[i]], cities[tmp[k + 1]]);
        }
    }

    return delta;
}


/*
    去重处理，对于Delta数组来说，对于城市序列1-2-3-4-5-6-7-8-9-10，如果对3-5应用了邻域操作2-opt ， 事实上对于
    7-10之间的翻转是不需要重复计算的。 所以用Delta提前预处理一下。

    当然由于这里的计算本身是O（1） 的，事实上并没有带来时间复杂度的减少（更新操作反而增加了复杂度） 
    如果delta计算 是O（n）的，这种去重操作效果是明显的。 
*/

void Update(int i, int k,  int *tmp, CITIES * cities){
    if (i && k != CITY_SIZE - 1){
        i --; k ++;
        for (int j = i; j <= k; j ++){
            for (int l = j + 1; l < CITY_SIZE; l ++){
                Delta[j][l] = calc_delta(j, l, tmp, cities);
            }
        }

        for (int j = 0; j < k; j ++){
            for (int l = i; l <= k; l ++){
                if (j >= l) continue;
                Delta[j][l] = calc_delta(j, l, tmp, cities);
            }
        }
    }// 如果不是边界，更新(i-1, k + 1)之间的 
    else{
        for (i = 0; i < CITY_SIZE - 1; i++)
            {
              for (k = i + 1; k < CITY_SIZE; k++)
            {
                Delta[i][k] = calc_delta(i, k, tmp, cities);
              }
            }  
    }// 边界要特殊更新 

} 

//本地局部搜索，边界条件 max_no_improve
//best_solution最优解
//current_solution当前解
void local_search(SOLUTION & best_solution, CITIES * cities, int max_no_improve)
{
    int count = 0;
    int i, k;

    int inital_cost = best_solution.cost; //初始花费

    int now_cost = 0;

    SOLUTION *current_solution = new SOLUTION; //为了防止爆栈……直接new了，你懂的

    for (i = 0; i < CITY_SIZE - 1; i++)
    {
       for (k = i + 1; k < CITY_SIZE; k++)
        {
            Delta[i][k] = calc_delta(i, k, best_solution.permutation, cities);
        }
    }

    do
    {
        //枚举排列
        for (i = 0; i < CITY_SIZE - 1; i++)
        {
            for (k = i + 1; k < CITY_SIZE; k++)
            {
                //邻域动作
                two_opt_swap(best_solution.permutation, current_solution->permutation, i, k);
                now_cost = inital_cost + Delta[i][k];
                current_solution->cost = now_cost;
                if (current_solution->cost < best_solution.cost)
                {
                    count = 0; //better cost found, so reset
                    for (int j = 0; j < CITY_SIZE; j++)
                    {
                        best_solution.permutation[j] = current_solution->permutation[j];
                    }
                    best_solution.cost = current_solution->cost;
                    inital_cost = best_solution.cost;
                    Update(i, k, best_solution.permutation, cities);
                }

            }
        }

        count++;

    } while (count <= max_no_improve);
}


//将城市序列分成4块，然后按块重新打乱顺序。
//用于扰动函数
void double_bridge_move(int *cities_permutation, int * new_cities_permutation)
{
    int temp_perm[CITY_SIZE];

    int pos1 = 1 + rand() % (CITY_SIZE / 4);
    int pos2 = pos1 + 1 + rand() % (CITY_SIZE / 4);
    int pos3 = pos2 + 1 + rand() % (CITY_SIZE / 4);

    int i;
    vector<int> v;
    //第一块
    for (i = 0; i < pos1; i++)
    {
        v.push_back(cities_permutation[i]);
    }

    //第二块
    for (i = pos3; i < CITY_SIZE; i++)
    {
        v.push_back(cities_permutation[i]);
    }
    //第三块
    for (i = pos2; i < pos3; i++)
    {
        v.push_back(cities_permutation[i]);
    }

    //第四块
    for (i = pos1; i < pos2; i++)
    {
        v.push_back(cities_permutation[i]);
    }


    for (i = 0; i < (int)v.size(); i++)
    {
        new_cities_permutation[i] = v[i];
    }    


}

//扰动
void perturbation(CITIES * cities, SOLUTION &best_solution, SOLUTION &current_solution)
{
    double_bridge_move(best_solution.permutation, current_solution.permutation);
    current_solution.cost = cost_total(current_solution.permutation, cities);
}

//迭代搜索
//max_iterations用于迭代搜索次数
//max_no_improve用于局部搜索边界条件
void iterated_local_search(SOLUTION & best_solution, CITIES * cities, int max_iterations, int max_no_improve)
{
    SOLUTION *current_solution = new SOLUTION;

    //获得初始随机解
    random_permutation(best_solution.permutation);


    best_solution.cost = cost_total(best_solution.permutation, cities);
    local_search(best_solution, cities, max_no_improve); //初始搜索

    for (int i = 0; i < max_iterations; i++)
    {
        perturbation(cities, best_solution, *current_solution); //扰动+判断是否接受新解
        local_search(*current_solution, cities, max_no_improve);//继续局部搜索

        //找到更优解
        if (current_solution->cost < best_solution.cost)
        {
            for (int j = 0; j < CITY_SIZE; j++)
            {
                best_solution.permutation[j] = current_solution->permutation[j];
            }
            best_solution.cost = current_solution->cost;
        }
        cout << setw(13) << setiosflags(ios::left) <<"迭代搜索 " << i << " 次\t" << "最优解 = " << best_solution.cost << " 当前解 = " << current_solution->cost << endl;
    }

}