TSP_旅行商問題-模擬退火演算法

• TSP_旅行商問題-貪心演算法
• TSP_旅行商問題-模擬退火演算法
• TSP_旅行商問題-遺傳演算法
• TSP_旅行商問題-基本蟻群演算法

問題描述

對於n組城市座標，尋找最短路徑使其經過所有城市並回到起點。

問題資料集：tsp.eil51問題

1 37 52
2 49 49
3 52 64
4 20 26
5 40 30
6 21 47
7 17 63
8 31 62
9 52 33
10 51 21
11 42 41
12 31 32
13 5 25
14 12 42
15 36 16
16 52 41
17 27 23
18 17 33
19 13 13
20 57 58
21 62 42
22 42 57
23 16 57
24 8 52
25 7 38
26 27 68
27 30 48
28 43 67
29 58 48
30 58 27
31 37 69
32 38 46
33 46 10
34 61 33
35 62 63
36 63 69
37 32 22
38 45 35
39 59 15
40 5 6
41 10 17
42 21 10
43 5 64
44 30 15
45 39 10
46 32 39
47 25 32
48 25 55
49 48 28
50 56 37
51 30 40
最優解：426
 

模擬退火演算法基本流程

1、設定初始溫度，隨機初始化一個初始解；2、開始進入外迴圈體
$\quad$2.a進入內迴圈體
$\quad$$\quad$2.a.a 根據一定規則，在現有解的基礎上生成一個新解；
$\quad$$\quad$2.a.b記錄內迴圈的最優解

$\quad$$\quad$2.a.c退出內迴圈
$\quad$2.b 按照一定規則，判斷是否接受新解，即使用新解替換舊解;
$\quad$2.c溫度下降
3、退出外迴圈

模擬退火演算法在tsp問題上的應用流程

1、設定初始溫度，隨機初始化一個初始解；2、開始進入外迴圈體

$$$\quad$2.a進入內迴圈體
$\quad$$\quad$2.a.a 在現有解即城市訪問序列中，隨機交換一組或多組城市的順序，以生成一個新解；
$\quad$$\quad$2.a.b記錄內迴圈的最優解
$\quad$$\quad$2.a.c退出內迴圈（終止條件我用的是迭代次數k）
$\quad$2.b 按照metropolis準則，根據新解的能量值（new_energy）、舊解的能量值（new_energy）和當前溫度 t 判斷是否接受新解，即使用新解替換舊解;
$\quad$2.c溫度下降（我設定的是接受新狀態，溫度下降的會快些，不接受會慢些降溫）
3、退出外迴圈（終止條件是溫度降低到一定程度）

【注】metropolis準則
$P=$$\begin{cases} 1,newenergy\quad lessthen\quad oldenergy;\\ \exp^{(-\frac{\Delta}{t})},newenergy\quad morethen\quad oldenergy;\\ \end{cases}$

$\Delta$指新舊解的能量差，是個正數
$P$指接受新解的概率

引數設定

初始溫度 tempterature = 1000；
成功降溫因子 u = 0.998；
失敗降溫因子 v = 0.999；
內迴圈迭代次數 k = 100;

測試結果

演算法程式碼

#include<iostream>
#include<ctime>
#include<cmath>
#include<fstream>
#include<algorithm>
using namespace std;
const int num = 1000;//city number
const int width = 100;
const int height = 100;

typedef struct node {
    int x;
    int y;
}city;
city citys[num];//citys
double dic[num][num];//distance from two citys;
bool visit[num];//visited
int N;//real citys
int seq[num];//最優路徑序列
double answer;//最優路徑長度
const int tempterature = 1000;//初始溫度
const double u = 0.998;//成功降溫因子
const double v = 0.999;//失敗降溫因子
int k = 100;//對每個溫度迭代次數
void init() {//set N&&x-y
    N = 51;
    citys[0].x = 37; citys[0].y = 52;
    citys[1].x = 49; citys[1].y = 49;
    citys[2].x = 52; citys[2].y = 64;
    citys[3].x = 20; citys[3].y = 26;
    citys[4].x = 40; citys[4].y = 30;
    citys[5].x = 21; citys[5].y = 47;
    citys[6].x = 17; citys[6].y = 63;
    citys[7].x = 31; citys[7].y = 62;
    citys[8].x = 52; citys[8].y = 33;
    citys[9].x = 51; citys[9].y = 21;
    citys[10].x = 42; citys[10].y = 41;
    citys[11].x = 31; citys[11].y = 32;
    citys[12].x = 5; citys[12].y = 25;
    citys[13].x = 12; citys[13].y = 42;
    citys[14].x = 36; citys[14].y = 16;
    citys[15].x = 52; citys[15].y = 41;
    citys[16].x = 27; citys[16].y = 23;
    citys[17].x = 17; citys[17].y = 33;
    citys[18].x = 13; citys[18].y = 13;
    citys[19].x = 57; citys[19].y = 58;
    citys[20].x = 62; citys[20].y = 42;
    citys[21].x = 42; citys[21].y = 57;
    citys[22].x = 16; citys[22].y = 57;
    citys[23].x = 8; citys[23].y = 52;
    citys[24].x = 7; citys[24].y = 38;
    citys[25].x = 27; citys[25].y = 68;
    citys[26].x = 30; citys[26].y = 48;
    citys[27].x = 43; citys[27].y = 67;
    citys[28].x = 58; citys[28].y = 48;
    citys[29].x = 58; citys[29].y = 27;
    citys[30].x = 37; citys[30].y = 69;
    citys[31].x = 38; citys[31].y = 46;
    citys[32].x = 46; citys[32].y = 10;
    citys[33].x = 61; citys[33].y = 33;
    citys[34].x = 62; citys[34].y = 63;
    citys[35].x = 63; citys[35].y = 69;
    citys[36].x = 32; citys[36].y = 22;
    citys[37].x = 45; citys[37].y = 35;
    citys[38].x = 59; citys[38].y = 15;
    citys[39].x = 5; citys[39].y = 6;
    citys[40].x = 10; citys[40].y = 17;
    citys[41].x = 21; citys[41].y = 10;
    citys[42].x = 5; citys[42].y = 64;
    citys[43].x = 30; citys[43].y = 15;
    citys[44].x = 39; citys[44].y = 10;
    citys[45].x = 32; citys[45].y = 39;
    citys[46].x = 25; citys[46].y = 32;
    citys[47].x = 25; citys[47].y = 55;
    citys[48].x = 48; citys[48].y = 28;
    citys[49].x = 56; citys[49].y = 37;
    citys[50].x = 30; citys[50].y = 40;
}
void set_dic() {//set distance
    for (int i = 0; i<N; ++i) {
        for (int j = 0; j<N; ++j) {
            dic[i][j] = sqrt(pow(citys[i].x - citys[j].x, 2) + pow(citys[i].y - citys[j].y, 2));
        }
    }
}
double dic_two_point(city a, city b) {
    return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));
}
double count_energy(int* conf) {
    double temp = 0;
    for (int i = 1; i<N; ++i) {
        temp += dic_two_point(citys[conf[i]], citys[conf[i - 1]]);
    }
    temp += dic_two_point(citys[conf[0]], citys[conf[N - 1]]);
    return temp;
}
bool metro(double f1, double f2, double t) {
    if (f2 < f1)
        return true;
    //else
    //  return false;
    double p = exp(-(f2 - f1) / t);
    int bignum = 1e9;
    if (rand() % bignum<p*bignum)
        return true;
    return false;
}
void generate(int* s) {//隨機產生一組新解
    bool v[num];
    memset(v, false, sizeof(v));
    for (int i = 0; i<N; ++i) {
        s[i] = rand() % N;
        while (v[s[i]]) {
            s[i] = rand() % N;
        }
        v[s[i]] = true;
    }
}
void generate1(int* s) {//隨機交換序列中的一組城市順序
    int ti = rand() % N;
    int tj = ti;
    while (ti == tj)
        tj = rand() % N;
    for (int i = 0; i<N; ++i)
        s[i] = seq[i];
    swap(s[ti], s[tj]);
}
void generate2(int* s) {//隨機交換序列中的兩組城市順序
    int ti = rand() % N;
    int tj = ti;
    int tk = ti;
    while (ti == tj)
        tj = rand() % N;
    while (ti == tj || tj == tk || ti == tk)
        tk = rand() % N;
    for (int i = 0; i<N; ++i)
        s[i] = seq[i];
    swap(s[ti], s[tj]);
    swap(s[tk], s[tj]);
}
void generate3(int* s) {//隨機選序列中的三個城市互相交換順序
    int ti = rand() % N;
    int tj = ti;
    int tm = rand() % N;
    int tn = ti;
    while (ti == tj)
        tj = rand() % N;
    while (tm == tn)
        tn = rand() % N;
    for (int i = 0; i<N; ++i)
        s[i] = seq[i];
    swap(s[ti], s[tj]);
    swap(s[tm], s[tn]);
}
void generate0(int* s) {//以上三種交換方式等概率選擇
    int temp = rand() % 3;
    if (temp == 0)
        generate1(s);
    else if (temp == 1)
        generate2(s);
    else if (temp == 2)
        generate3(s);
}
void moni() {
    double t = tempterature;
    int seq_t[num];
    for (int i = 0; i<N; ++i) {//初始化當前序列
        seq[i] = seq_t[i] = i;
    }
    double new_energy = 1, old_energy = 0;
    while (t>1e-9&&fabs(new_energy - old_energy)>1e-9) {//溫度作為控制變數
        int t_k = k;
        int seq_tt[num];
        while (t_k--&&fabs(new_energy - old_energy)>1e-9) {//迭代次數作為控制變數
            generate1(seq_tt);
            new_energy = count_energy(seq_tt);//new
            old_energy = count_energy(seq_t);//old
            if (metro(old_energy, new_energy, t))
                for (int i = 0; i < N; ++i)
                    seq_t[i] = seq_tt[i];
        }
        new_energy = count_energy(seq_t);//new
        old_energy = answer;//old
        if (metro(old_energy, new_energy, t)) {
            for (int i = 0; i < N; ++i)
                seq[i] = seq_t[i];
            answer = count_energy(seq);
            t *= u;//接受新狀態降溫因子0.98
        }
        else
            t *= v;//不接受新狀態降溫因子0.99
    }
    answer = count_energy(seq);
}
void output() {
    cout << "the best road is : \n";
    for (int i = 0; i < N; ++i) {
        cout << seq[i];
        if (i == N - 1)
            cout << endl;
        else
            cout << " -> ";
    }
    cout << "the length of the road is " << answer << endl;
}
void test() {
    ifstream ifile("data.txt");
    if (!ifile) {
        cout << "open field\n";
        return;
    }
    while (!ifile.eof()) {
        int te = 0;
        ifile >> te;
        ifile >> citys[te - 1].x >> citys[te - 1].y;
    }
}
int main() {
    srand(time(nullptr));
    int t;
    while (cin >> t) {//僅作為重啟演算法開關使用，無意義
        init();//使用程式內建資料使用init()函式，
        //test();//使用檔案讀取資料使用test()函式，
        set_dic();//計算每個城市之間的距離
        moni();//退火
        output();//輸出
    }
    return 0;
}