【优化算法】迭代局部搜索算法(Iterated local search)探幽（附C++代码及注释）

  1////////////////////////
  2//TSP问题 迭代局部搜索求解代码
  3//基于Berlin52例子求解
  4//作者：infinitor
  5//时间：2018-04-12
  6////////////////////////
  7
  8
  9#include <iostream>
 10#include <cmath>
 11#include <stdlib.h>
 12#include <time.h>
 13#include <vector>
 14#include <windows.h>
 15#include <memory.h>
 16#include <string.h>
 17#include <iomanip>
 18
 19#define DEBUG
 20
 21using namespace std;
 22
 23#define CITY_SIZE 52 //城市数量
 24
 25
 26//城市坐标
 27typedef struct candidate
 28{
 29    int x;
 30    int y;
 31}city, CITIES;
 32
 33//优化值
 34int **Delta; 
 35
 36//解决方案
 37typedef struct Solution
 38{
 39    int permutation[CITY_SIZE]; //城市排列
 40    int cost;                        //该排列对应的总路线长度
 41}SOLUTION;
 42// 计算邻域操作优化值 
 43int calc_delta(int i, int k, int *tmp, CITIES * cities);
 44
 45//计算两个城市间距离
 46int distance_2city(city c1, city c2);
 47
 48//根据产生的城市序列，计算旅游总距离
 49int cost_total(int * cities_permutation, CITIES * cities);
 50
 51//获取随机城市排列, 用于产生初始解
 52void random_permutation(int * cities_permutation);
 53
 54//颠倒数组中下标begin到end的元素位置, 用于two_opt邻域动作
 55void swap_element(int *p, int begin, int end);
 56
 57//邻域动作 反转index_i <-> index_j 间的元素
 58void two_opt_swap(int *cities_permutation, int *new_cities_permutation, int index_i, int index_j);
 59
 60//本地局部搜索，边界条件 max_no_improve
 61void local_search(SOLUTION & best, CITIES * cities, int max_no_improve);
 62
 63//判断接受准则
 64bool AcceptanceCriterion(int *cities_permutation, int *old_cities_permutation, CITIES * p_cities);
 65
 66//将城市序列分成4块，然后按块重新打乱顺序。
 67//用于扰动函数
 68void double_bridge_move(int *cities_permutation, int * new_cities_permutation);
 69
 70//扰动
 71void perturbation(CITIES * cities, SOLUTION &best_solution, SOLUTION &current_solution);
 72
 73//迭代搜索
 74void iterated_local_search(SOLUTION & best, CITIES * cities, int max_iterations, int max_no_improve);
 75
 76// 更新Delta 
 77void Update(int i, int k,  int *tmp, CITIES * cities);
 78
 79//城市排列
 80int permutation[CITY_SIZE];
 81//城市坐标数组
 82CITIES cities[CITY_SIZE];
 83
 84
 85//berlin52城市坐标，最优解7542好像
 86CITIES berlin52[CITY_SIZE] = { { 565,575 },{ 25,185 },{ 345,750 },{ 945,685 },{ 845,655 },
 87{ 880,660 },{ 25,230 },{ 525,1000 },{ 580,1175 },{ 650,1130 },{ 1605,620 },
 88{ 1220,580 },{ 1465,200 },{ 1530,5 },{ 845,680 },{ 725,370 },{ 145,665 },
 89{ 415,635 },{ 510,875 },{ 560,365 },{ 300,465 },{ 520,585 },{ 480,415 },
 90{ 835,625 },{ 975,580 },{ 1215,245 },{ 1320,315 },{ 1250,400 },{ 660,180 },
 91{ 410,250 },{ 420,555 },{ 575,665 },{ 1150,1160 },{ 700,580 },{ 685,595 },
 92{ 685,610 },{ 770,610 },{ 795,645 },{ 720,635 },{ 760,650 },{ 475,960 },
 93{ 95,260 },{ 875,920 },{ 700,500 },{ 555,815 },{ 830,485 },{ 1170,65 },
 94{ 830,610 },{ 605,625 },{ 595,360 },{ 1340,725 },{ 1740,245 } };
 95
 96int main()
 97{
 98    srand(1);
 99    int max_iterations = 600;
100    int max_no_improve = 50;
101    //初始化指针数组 
102    Delta = new int*[CITY_SIZE];
103    for (int i = 0; i < CITY_SIZE; i ++)
104        Delta[i] = new int[CITY_SIZE];
105
106    SOLUTION best_solution;
107
108    iterated_local_search(best_solution, berlin52, max_iterations, max_no_improve);
109
110    cout << endl<<endl<<"搜索完成！ 最优路线总长度 = " << best_solution.cost << endl;
111    cout << "最优访问城市序列如下：" << endl;
112    for (int i = 0; i < CITY_SIZE;i++)
113    {
114        cout << setw(4) << setiosflags(ios::left) << best_solution.permutation[i];
115    }
116
117    cout << endl << endl;
118
119    return 0;
120}
121
122
123
124//计算两个城市间距离
125int distance_2city(city c1, city c2)
126{
127    int distance = 0;
128    distance = sqrt((double)((c1.x - c2.x)*(c1.x - c2.x) + (c1.y - c2.y)*(c1.y - c2.y)));
129
130    return distance;
131}
132
133//根据产生的城市序列，计算旅游总距离
134//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
135//访问顺序不同，那么总路线长度也是不同的
136//p_perm 城市序列参数
137int cost_total(int * cities_permutation, CITIES * cities)
138{
139    int total_distance = 0;
140    int c1, c2;
141    //逛一圈，看看最后的总距离是多少
142    for (int i = 0; i < CITY_SIZE; i++)
143    {
144        c1 = cities_permutation[i];
145        if (i == CITY_SIZE - 1) //最后一个城市和第一个城市计算距离
146        {
147            c2 = cities_permutation[0];
148        }
149        else
150        {
151            c2 = cities_permutation[i + 1];
152        }
153        total_distance += distance_2city(cities[c1], cities[c2]);
154    }
155
156    return total_distance;
157}
158
159//获取随机城市排列
160void random_permutation(int * cities_permutation)
161{
162    int i, r, temp;
163    for (i = 0; i < CITY_SIZE; i++)
164    {
165        cities_permutation[i] = i; //初始化城市排列，初始按顺序排
166    }
167
168
169    for (i = 0; i < CITY_SIZE; i++)
170    {
171        //城市排列顺序随机打乱
172        r = rand() % (CITY_SIZE - i) + i;
173        temp = cities_permutation[i];
174        cities_permutation[i] = cities_permutation[r];
175        cities_permutation[r] = temp;
176    }
177}
178
179
180
181
182//颠倒数组中下标begin到end的元素位置
183void swap_element(int *p, int begin, int end)
184{
185    int temp;
186    while (begin < end)
187    {
188        temp = p[begin];
189        p[begin] = p[end];
190        p[end] = temp;
191        begin++;
192        end--;
193    }
194}
195
196
197//邻域动作 反转index_i <-> index_j 间的元素
198void two_opt_swap(int *cities_permutation, int *new_cities_permutation, int index_i, int index_j)
199{
200    for (int i = 0; i < CITY_SIZE; i++)
201    {
202        new_cities_permutation[i] = cities_permutation[i];
203    }
204
205    swap_element(new_cities_permutation, index_i, index_j);
206}
207
208
209
210int calc_delta(int i, int k,  int *tmp, CITIES * cities){
211    int delta = 0;
212    /*
213                以下计算说明：
214                对于每个方案，翻转以后没必要再次重新计算总距离
215                只需要在翻转的头尾做个小小处理
216
217                比如：
218                有城市序列   1-2-3-4-5 总距离 = d12 + d23 + d34 + d45 + d51 = A
219                翻转后的序列 1-4-3-2-5 总距离 = d14 + d43 + d32 + d25 + d51 = B
220                由于 dij 与 dji是一样的，所以B也可以表示成 B = A - d12 - d45 + d14 + d25
221                下面的优化就是基于这种原理
222    */
223    if (i == 0)
224    {
225        if (k == CITY_SIZE - 1)
226        {
227           delta = 0;
228        }
229        else
230        {
231            delta = 0
232                - distance_2city(cities[tmp[k]], cities[tmp[k + 1]])
233                + distance_2city(cities[tmp[i]], cities[tmp[k + 1]])
234                - distance_2city(cities[tmp[CITY_SIZE - 1]], cities[tmp[i]])
235                + distance_2city(cities[tmp[CITY_SIZE - 1]], cities[tmp[k]]);
236        }
237
238    }
239    else
240    {
241        if (k == CITY_SIZE - 1)
242        {
243            delta = 0
244                - distance_2city(cities[tmp[i - 1]], cities[tmp[i]])
245                + distance_2city(cities[tmp[i - 1]], cities[tmp[k]])
246                - distance_2city(cities[tmp[0]], cities[tmp[k]])
247                + distance_2city(cities[tmp[i]], cities[tmp[0]]);
248        }
249        else
250        {
251            delta = 0
252                - distance_2city(cities[tmp[i - 1]], cities[tmp[i]])
253                + distance_2city(cities[tmp[i - 1]], cities[tmp[k]])
254                - distance_2city(cities[tmp[k]], cities[tmp[k + 1]])
255                + distance_2city(cities[tmp[i]], cities[tmp[k + 1]]);
256        }
257    }
258
259    return delta;
260}
261
262
263/*
264    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
265    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。
266
267    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） 
268    如果delta计算 是O（n）的，这种去重操作效果是明显的。 
269*/
270
271void Update(int i, int k,  int *tmp, CITIES * cities){
272    if (i && k != CITY_SIZE - 1){
273        i --; k ++;
274        for (int j = i; j <= k; j ++){
275            for (int l = j + 1; l < CITY_SIZE; l ++){
276                Delta[j][l] = calc_delta(j, l, tmp, cities);
277            }
278        }
279
280        for (int j = 0; j < k; j ++){
281            for (int l = i; l <= k; l ++){
282                if (j >= l) continue;
283                Delta[j][l] = calc_delta(j, l, tmp, cities);
284            }
285        }
286    }// 如果不是边界，更新(i-1, k + 1)之间的 
287    else{
288        for (i = 0; i < CITY_SIZE - 1; i++)
289            {
290              for (k = i + 1; k < CITY_SIZE; k++)
291            {
292                Delta[i][k] = calc_delta(i, k, tmp, cities);
293              }
294            }  
295    }// 边界要特殊更新 
296
297} 
298
299//本地局部搜索，边界条件 max_no_improve
300//best_solution最优解
301//current_solution当前解
302void local_search(SOLUTION & best_solution, CITIES * cities, int max_no_improve)
303{
304    int count = 0;
305    int i, k;
306
307    int inital_cost = best_solution.cost; //初始花费
308
309    int now_cost = 0;
310
311    SOLUTION *current_solution = new SOLUTION; //为了防止爆栈……直接new了，你懂的
312
313    for (i = 0; i < CITY_SIZE - 1; i++)
314    {
315       for (k = i + 1; k < CITY_SIZE; k++)
316        {
317            Delta[i][k] = calc_delta(i, k, best_solution.permutation, cities);
318        }
319    }
320
321    do
322    {
323        //枚举排列
324        for (i = 0; i < CITY_SIZE - 1; i++)
325        {
326            for (k = i + 1; k < CITY_SIZE; k++)
327            {
328                //邻域动作
329                two_opt_swap(best_solution.permutation, current_solution->permutation, i, k);
330                now_cost = inital_cost + Delta[i][k];
331                current_solution->cost = now_cost;
332                if (current_solution->cost < best_solution.cost)
333                {
334                    count = 0; //better cost found, so reset
335                    for (int j = 0; j < CITY_SIZE; j++)
336                    {
337                        best_solution.permutation[j] = current_solution->permutation[j];
338                    }
339                    best_solution.cost = current_solution->cost;
340                    inital_cost = best_solution.cost;
341                    Update(i, k, best_solution.permutation, cities);
342                }
343
344            }
345        }
346
347        count++;
348
349    } while (count <= max_no_improve);
350}
351//判断接受准则
352bool AcceptanceCriterion(int *cities_permutation, int *old_cities_permutation, CITIES * p_cities)
353{
354    int acceptance = 500; //接受条件,与当前最解相差不超过acceptance
355    int old_cost = cost_total(old_cities_permutation, p_cities);
356    int new_cost = cost_total(cities_permutation, p_cities);
357
358    if ((new_cost <= (old_cost + acceptance)) || (new_cost >= (old_cost - acceptance)))
359    {
360        return true;
361    }
362
363    return false;
364}
365
366//将城市序列分成4块，然后按块重新打乱顺序。
367//用于扰动函数
368void double_bridge_move(int *cities_permutation, int * new_cities_permutation)
369{
370    int temp_perm[CITY_SIZE];
371
372    int pos1 = 1 + rand() % (CITY_SIZE / 4);
373    int pos2 = pos1 + 1 + rand() % (CITY_SIZE / 4);
374    int pos3 = pos2 + 1 + rand() % (CITY_SIZE / 4);
375
376    int i;
377    vector<int> v;
378    //第一块
379    for (i = 0; i < pos1; i++)
380    {
381        v.push_back(cities_permutation[i]);
382    }
383
384    //第二块
385    for (i = pos3; i < CITY_SIZE; i++)
386    {
387        v.push_back(cities_permutation[i]);
388    }
389    //第三块
390    for (i = pos2; i < pos3; i++)
391    {
392        v.push_back(cities_permutation[i]);
393    }
394
395    //第四块
396    for (i = pos1; i < pos2; i++)
397    {
398        v.push_back(cities_permutation[i]);
399    }
400
401
402    for (i = 0; i < (int)v.size(); i++)
403    {
404        temp_perm[i] = v[i];
405    }
406    //if accept判断是否接受当前解
407    if (AcceptanceCriterion(cities_permutation, temp_perm, cities))
408    {
409        memcpy(new_cities_permutation, temp_perm, sizeof(temp_perm));//accept
410    }
411
412
413}
414
415//扰动
416void perturbation(CITIES * cities, SOLUTION &best_solution, SOLUTION &current_solution)
417{
418    double_bridge_move(best_solution.permutation, current_solution.permutation);
419    current_solution.cost = cost_total(current_solution.permutation, cities);
420}
421
422//迭代搜索
423//max_iterations用于迭代搜索次数
424//max_no_improve用于局部搜索边界条件
425void iterated_local_search(SOLUTION & best_solution, CITIES * cities, int max_iterations, int max_no_improve)
426{
427    SOLUTION *current_solution = new SOLUTION;
428
429    //获得初始随机解
430    random_permutation(best_solution.permutation);
431
432
433    best_solution.cost = cost_total(best_solution.permutation, cities);
434    local_search(best_solution, cities, max_no_improve); //初始搜索
435
436    for (int i = 0; i < max_iterations; i++)
437    {
438        perturbation(cities, best_solution, *current_solution); //扰动+判断是否接受新解
439        local_search(*current_solution, cities, max_no_improve);//继续局部搜索
440
441        //找到更优解
442        if (current_solution->cost < best_solution.cost)
443        {
444            for (int j = 0; j < CITY_SIZE; j++)
445            {
446                best_solution.permutation[j] = current_solution->permutation[j];
447            }
448            best_solution.cost = current_solution->cost;
449        }
450        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;
451    }
452
453}