TSP_旅行商問題 - 貪心演算法

• 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

尋找最短路徑使得其經過所有城市
測試資料集：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

演算法思想

選擇下一城市的策略為
距當前城市最近且未被訪問過的城市

演算法流程

準備工作（初始化）

a、讀取資料，txt內資料格式為：序號 $x_i$ $y_i$
b、設定城市數量N、城市座標陣列citys[num]
c、計算城市距離矩陣，

開始模擬演算法流程
seq[num]記錄路徑，
visit[num]標記是否已訪問，

a、初始化visit陣列為false未訪問；
b、隨機選擇起點城市，記錄路徑，標記visit[seq[0]]為true已訪問；

進入迴圈體
迴圈N-1次

a、遍歷所有城市，尋找未訪問且與上一城市seq[i-1]距離最近的城市mini；
b、記錄路徑，標記visit[mini]為true已訪問；

結束迴圈

計算路徑能量值

測試結果

當前結果

最優結果

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];//
double answer;
void init() {//set N&&x-y設定N和citys[num]
    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;
}
void moni() {
    memset(visit, false, sizeof(visit));
    int temp = rand() % N;
    seq[0] = temp;
    visit[temp] = true;
    int mini = -1;
    int ans = 1e9;
    for (int i = 1; i < N; ++i) {//第i位應該經過的點
        ans = 1e9;
        mini = -1;
        for (int j = 0; j < N; ++j) {
            if (!visit[j] && ans > dic[seq[i - 1]][j]) {
                ans = dic[seq[i - 1]][j];
                mini = j;
            }
        }
        seq[i] = mini;
        visit[mini] = true;
    }
    answer=count_energy(seq);
}
void test() {//讀取資料，設定N和citys[num]
    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;
        N = te;
    }
}
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;
}
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();
        moni();
        output();
    }
    return 0;
}