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TSP_旅行商问题 - 贪心算法
问题描述
寻找最短路径使得其经过所有城市
测试数据集:tsp.eil51问题
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
最优解:426
算法思想
选择下一城市的策略为
距当前城市最近且未被访问过的城市
算法流程
准备工作(初始化)
a、读取数据,txt内数据格式为:序号
b、设置城市数量N、城市坐标数组citys[num]
c、计算城市距离矩阵,
为第i个城市到第j个城市的距离
开始模拟算法流程
seq[num]记录路径,
visit[num]标记是否已访问,
a、初始化visit数组为false未访问;
b、随机选择起点城市,记录路径,标记visit[seq[0]]为true已访问;
进入循环体
循环N-1次
a、遍历所有城市,寻找未访问且与上一城市seq[i-1]距离最近的城市mini;
b、记录路径,标记visit[mini]为true已访问;
结束循环
计算路径能量值
测试结果
当前结果
最优结果
算法代码
#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设置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位应该经过的点
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() {//读取数据,设置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) {//仅作为重启算法开关使用,无意义
init();//使用程序内置数据使用init()函数,
//test();//使用文件读取数据使用test()函数,
set_dic();
moni();
output();
}
return 0;
}