TSP_旅行商問題 - 貪心演算法
阿新 • • 發佈:2018-12-25
TSP_旅行商問題 - 貪心演算法
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
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;
}