分布估计算法解决旅行商问题(TSP)
TSP问题(Traveling Salesman Problem,旅行商问题),由威廉哈密顿爵士和英国数学家克克曼T.P.Kirkman于19世纪初提出。问题描述如下:
有若干个城市,任何两个城市之间的距离都是确定的,现要求一旅行商从某城市出发必须经过每一个城市且只在一个城市逗留一次,最后回到出发的城市,问如何事先确定一条最短的线路已保证其旅行的费用最少?
下面采用分布估计算法来解决旅行商问题。
在用分布估计算法解决旅行商问题时,结构与传统的分布估计算法相似,只是把概率向量换位概率矩阵而已:
1. 通过概率矩阵生成解。
2. 评估解。
3. 选择优势群体。
4. 更新概率矩阵。
5. 重复以上4步直至迭代结束。
这里说的“概率矩阵”记录了上一代优势群体中,“城市对”出现的次数(或与城市对出现次数成正比的一个数)。“城市对”是指路线中相邻的两个城市,这两个城市不分先后。假设概率矩阵第i行第j列的元素为 pijp_{ij}pij,它代表在优势群体中城市i和城市j相邻的次数(即i−j和j−i发生的次数)。
经过测试(于2019-07-09),这种算法非常容易陷入局部最优,效果不是很好。
具体实现:
C++:
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int city_num = 10;//城市数量
const int unit_num = 100;//群体规模
int ps = 10;//变异概率
const int genmax = 500;//最大迭代数
const int unit_num = 100;//群体规模
int ps = 10;//变异概率
const int genmax = 500;//最大迭代数
//城市间距离映射 最优解权值=10
int length_table[10][10] = {
int length_table[10][10] = {
{0,1,1272,2567,1653,2097,1425,1177,3947,1},
{1,0,1,2511,1633,2077,1369,1157,3961,1518},
{1272,1,0,1,380,1490,821,856,3660,385},
{2567,2511,1,0,1,2335,1562,2165,3995,933},
{1653,1633,380,1,0,1,1041,1135,3870,456},
{2097,2077,1490,2335,1,0,1,920,2170,1920},
{1425,1369,821,1562,1041,1,0,1,4290,626},
{1177,1157,856,2165,1135,920,1,0,1,1290},
{3947,3961,3660,3995,3870,2170,4290,1,0,1},
{1,1518,385,993,456,1920,626,1290,1,0}
};
};
class Unit
{
public:
int path[city_num];//个体的路径信息
int length;//个体价值
};
{
public:
int path[city_num];//个体的路径信息
int length;//个体价值
};
class Group
{
public:
Unit group[unit_num];
Unit best;
int best_gen;
{
public:
Unit group[unit_num];
Unit best;
int best_gen;
Group()
{
best.length = 0x3f3f3f3f;
best_gen = 0;
for(int i = 0; i < unit_num; i++)
{
bool flag[city_num] = {};
{
best.length = 0x3f3f3f3f;
best_gen = 0;
for(int i = 0; i < unit_num; i++)
{
bool flag[city_num] = {};
for(int j = 0; j < city_num; j++)
{
int t_city = rand()%city_num;
while(flag[t_city])
t_city = rand()%city_num;
flag[t_city] = true;
group[i].path[j] = t_city;
}
}
}
{
int t_city = rand()%city_num;
while(flag[t_city])
t_city = rand()%city_num;
flag[t_city] = true;
group[i].path[j] = t_city;
}
}
}
//对每个个体进行评估
void assess()
{
for(int k = 0; k < unit_num; k++)
{
int rel = 0;
for(int i = 1; i < city_num; i++)
rel += length_table[group[k].path[i-1]][group[k].path[i]];
rel += length_table[group[k].path[city_num-1]][group[k].path[0]];
group[k].length = rel;
}
}
void assess()
{
for(int k = 0; k < unit_num; k++)
{
int rel = 0;
for(int i = 1; i < city_num; i++)
rel += length_table[group[k].path[i-1]][group[k].path[i]];
rel += length_table[group[k].path[city_num-1]][group[k].path[0]];
group[k].length = rel;
}
}
//根据评估结果对个体进行排序
void unit_sort()
{
for(int i = 0; i < unit_num; i++)
{
for(int j = i+1; j < unit_num; j++)
{
if(group[i].length > group[j].length)
{
Unit temp;
memcpy(&temp, &group[i], sizeof(Unit));
memcpy(&group[i], &group[j], sizeof(Unit));
memcpy(&group[j], &temp, sizeof(Unit));
}
}
}
}
void unit_sort()
{
for(int i = 0; i < unit_num; i++)
{
for(int j = i+1; j < unit_num; j++)
{
if(group[i].length > group[j].length)
{
Unit temp;
memcpy(&temp, &group[i], sizeof(Unit));
memcpy(&group[i], &group[j], sizeof(Unit));
memcpy(&group[j], &temp, sizeof(Unit));
}
}
}
}
//交叉
Unit cross(Unit &father, Unit &mother)
{
int l = rand()%city_num;
int r = rand()%city_num;
if(l > r)
swap(l, r);
Unit cross(Unit &father, Unit &mother)
{
int l = rand()%city_num;
int r = rand()%city_num;
if(l > r)
swap(l, r);
bool flag[city_num] = {};
for(int i = l; i <= r; i++)
flag[father.path[i]] = true;
for(int i = l; i <= r; i++)
flag[father.path[i]] = true;
Unit son;
int pos = 0;
int pos = 0;
for(int i = 0; i < l; i++)
{
while(flag[mother.path[pos]])
pos++;
son.path[i] = mother.path[pos++];
}
for(int i = l; i <= r; i++)
son.path[i] = father.path[i];
for(int i = r+1; i < city_num; i++)
{
while(flag[mother.path[pos]])
pos++;
son.path[i] = mother.path[pos++];
}
{
while(flag[mother.path[pos]])
pos++;
son.path[i] = mother.path[pos++];
}
for(int i = l; i <= r; i++)
son.path[i] = father.path[i];
for(int i = r+1; i < city_num; i++)
{
while(flag[mother.path[pos]])
pos++;
son.path[i] = mother.path[pos++];
}
return son;
}
}
//突变
void mutation(Unit &t)
{
int proport = rand() % 100;
void mutation(Unit &t)
{
int proport = rand() % 100;
if(proport > ps)
return;
int one = rand()%city_num;
int two = rand()%city_num;
while(two != one)
two = rand()%city_num;
swap(t.path[one], t.path[two]);
}
return;
int one = rand()%city_num;
int two = rand()%city_num;
while(two != one)
two = rand()%city_num;
swap(t.path[one], t.path[two]);
}
//输出信息
void print()
{
for(int i = 0; i < unit_num; i++)
{
printf("第%d个个体,路径信息:", i);
for(int j = 0; j < city_num; j++)
printf("%d ", group[i].path[j]);
void print()
{
for(int i = 0; i < unit_num; i++)
{
printf("第%d个个体,路径信息:", i);
for(int j = 0; j < city_num; j++)
printf("%d ", group[i].path[j]);
printf(";总权值:%d;\n", group[i].length);
}
printf("最优个体,路径信息:");
for(int j = 0; j < city_num; j++)
printf("%d ", group[0].path[j]);
}
printf("最优个体,路径信息:");
for(int j = 0; j < city_num; j++)
printf("%d ", group[0].path[j]);
printf(";总权值:%d;\n", group[0].length);
}
}
//种群进化
void work()
{
for(int i = 0; i < genmax; i++)
{
//如果进化层数大于20,加大变异的概率
if(i > 20)
ps *= 3;
void work()
{
for(int i = 0; i < genmax; i++)
{
//如果进化层数大于20,加大变异的概率
if(i > 20)
ps *= 3;
assess();//评估
unit_sort();//根据评估结果排序
if(best.length > group[0].length)
{
memcpy(&best, &group[0], sizeof(group[0]));
best_gen = i;
}
{
memcpy(&best, &group[0], sizeof(group[0]));
best_gen = i;
}
for(int j = 0; j+2 < unit_num; j+=3)
group[j+2] = cross(group[j], group[j+1]);
group[j+2] = cross(group[j], group[j+1]);
for(int j = 0; j < city_num; j++)//变异(从1开始,保留最优)
mutation(group[j]);
}
}
mutation(group[j]);
}
}
};
Unit group[unit_num];//种群变量
Unit bestone;//记录最短路径
int generation_num;//记录当前达到了第几代
Unit bestone;//记录最短路径
int generation_num;//记录当前达到了第几代
int main()
{
srand((int)time(0));
{
srand((int)time(0));
for(int i = 0; i < 20; i++)
{
Group g;
g.work();
printf("第%d次求解。路径:", i+1);
for(int j = 0; j < city_num; j++)
printf("%d ", g.best.path[j]);
{
Group g;
g.work();
printf("第%d次求解。路径:", i+1);
for(int j = 0; j < city_num; j++)
printf("%d ", g.best.path[j]);
printf(";总权值:%d; 第%d代;\n", g.best.length, g.best_gen);
}
return 0;
}
}
return 0;
}