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发表于 2010-8-17 20:42:23
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来自 湖南长沙
此题的最终做的结果和大家分享一下:
首先是自己设计了一个算法,不能求得最优解,但是能够搜索到较为接近最优解的解。matlab代码如下:
- function resem_prim(Q,n,order)
- %参数及符号说明:
- %Q:邻接矩阵
- %n:矩阵的大小
- %order:地势的排序矩阵
- %index:记录迭代进入的层数
- %QQ:最小树的邻接矩阵
- %cost:最小生成路径的花费
- %result:生成树节点矩阵
- %weight:该点的流量/100
- %%%%%%%%%%%%%%%%%%%%%%%%%%主程序%%%%%%%%%%%%%%%%%%%%%%%%%%%
- %% 矩阵处理,按地势排序
- %此例的顺序为8,5,(4,10),7,3,(1,9)2,6
- for i=1:n
- for j=1:n
- temp(i,j)=Q(order(i),order(j));
- end
- end
- Q=temp;
- %% 搜索初始化
- QQ=[inf,Q(1,2);Q(2,1),inf];
- %cost=QQ(1,2)*100^0.51*0.66;
- weight=zeros(1,size(order,2));
- weight(1)=2;%该
- weight(2)=1;
- %% 逐层搜索
- for i=3:n
- weight(i)=weight(i)+1;
- [QQ,cost,weight]=compare(QQ,Q,i,weight);
-
- % QQ(i,:)=inf;
- % QQ(:,i)=inf;
- % QQ(match,i)=Q(match,i);
- % QQ(i,match)=Q(match,i);
- end
- %cost=cost+100.^0.51*0.66*QQ(10,1);
- %% 结果输出
- disp('经过搜索,得出最优的路径如下:');
- for i=1:n
- for j=i+1:n
- if QQ(i,j)~=inf
- disp([num2str(order(i)) '->' num2str(order(j))]);
- end
- end
- end
- disp('经计算得,最优花费为:');
- disp(cost);
- QQ
- %%%%%%%%%%%%%%%%%%%%%%%%%%主程序%%%%%%%%%%%%%%%%%%%%%%%%%%%
- %% 寻优函数
- function [QQ,min_cost,weight]=compare(QQ,Q,i,weight)
- min_cost=inf;
- test_QQ=QQ;
- test_weight=weight;
- for j=1:i-1
- test_QQ(i,j)=Q(i,j);
- test_QQ(j,i)=Q(j,i);
- test_QQ=fulllize(test_QQ);
- test_weight=change(test_weight,j,test_QQ);
- if cost(test_QQ,test_weight)<min_cost
- min_cost=cost(test_QQ,weight);
- flag=j;
- end
- test_QQ=QQ;
- test_weight=weight;
- end
- weight=change(weight,flag,QQ);
- QQ(i,flag)=Q(i,flag);
- QQ(flag,i)=Q(flag,i);
- QQ=fulllize(QQ);
- %% 计算花费的函数
- function cost=cost(QQ,weight)
- cost=0;
- for i=1:size(QQ,2)
- for j=i+1:size(QQ,2)
- if QQ(i,j)~=inf
- cost=cost+(weight(j)*100).^0.51*0.66*QQ(i,j);
- end
- end
- end
- %% 处理矩阵的函数
- function QQ=fulllize(QQ)
- for i=1:size(QQ,1)
- for j=1:size(QQ,2)
- if QQ(i,j)==0
- QQ(i,j)=inf;
- end
- end
- end
- %% 改变矩阵
- function weight=change(weight,j,QQ)
- weight(j)=weight(j)+1;
- for i=j-1:-1:1
- if(QQ(j,i)~=inf)
- weight=change(weight,i,QQ);
- end
- end
复制代码
还尝试利用lingo11.0进行求解,能够求解得到全局最优解,不过求解速度比较慢,不适合作为一个普适的算法,lingo代码如下:
- sets:
- site/1..10/:m;
- links(site,site):x,L;
- endsets
- data:
- L= 0 8 8 14 11 9 16 17 8 14
- 8 0 14 8 9 11 22 15 17 18
- 8 14 0 17 12 6 10 15 15 11
- 14 8 17 0 5 12 22 9 12 16
- 11 9 12 5 0 7 17 7 9 12
- 9 11 6 12 7 0 11 10 10 8
- 16 22 10 22 17 11 0 18 15 11
- 17 15 15 9 7 10 18 0 3 7
- 8 17 15 12 9 10 15 3 0 6
- 14 18 11 16 12 8 11 7 6 0;
- enddata
- [OBJ]min=0.66*@sum(site(i)|i#Le#9:@sum(site(j)|j#GT#i:x(i,j)^0.51*L(i,j))*m(i)^0.51);
- !约束一;
- m(1)=100;
- @for(site(i)|i#Le#9 #and# i#ge#2:m(i)-100=@sum(site(k)|k#le#i-1:x(k,i)*m(k)));
- @for(site(i)|i#le#9:@sum(site(j)|j#gt#i:x(i,j))>=1);
- @for(site(i)|i#le#9:@for(site(j)|j#le#10 #and# j#gt#i:@bin(x(i,j))));
复制代码
前种算法搜不到精确的最优解,后种方法效率不高,要是村庄再多一点的话要算很长的时间。
不知道大家有什么意见? |
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