【漫漫科研路\pgfplots】最小跳数最大权重算法

上周，实验室国际友人让我帮忙实现满足条件的最小跳数最大权重的算法。他的具体问题如下：
给定一个权重图（如下图所示），给出节点之间最小跳数最大权重矩阵，其中任意两点之间跳数小于等于$3$，否则权重为inf。

如图1所示， A到F的最小跳数为$2$：$A-C-F$和$A-E-F$，权重(这里权重表示为所有路径上的权重乘积，当然也可以稍加修改变成权重和)分别为$4\times1=4$、$3\times 4=12$。因此$A$到$F$的最小跳数最大权重为$12$，路径为$A-E-F$。下面给出了具体的代码实现：
主要有两个文件，测试脚本文件main.m和dijkstra_all.m函数文件：
1、测试脚本文件main.m

clear all
clc   
AdjMatrix=[0 inf 4 6 3 inf;
           inf 0 3 2 inf 4;
           4 3 0 1 1 1;
           6 2 1 0 inf inf;
           3 inf 1 inf 0 4;
           inf 4 1 inf 4 0;];
       
AdjMatrix1=AdjMatrix;% weight matrix
IND=AdjMatrix<inf&AdjMatrix>0;
AdjMatrix(IND)=1;% adjacent matrix
       
       
ResMatrix=zeros(size(AdjMatrix));%ouput matrix: the weights between each pair of nodes
N=length(AdjMatrix);% the number of nodes


for i=1:N
    for j=1:N
        if(i==j)
            ResMatrix(i,j)=0;
        else
            [sp, spcost]=dijkstra_all(AdjMatrix, i, j);% condition 1: find all the minimum hops    
            temp_num=sum(sp(1,:)>0);
            if(temp_num<=4)% condition 2: the number of the minimum hop is less than 3
                temp=ones(1,size(sp,1));% the number of the minimum hops
                for m=1:size(sp,1)
                    for k=1:temp_num-1
                        temp(m)=temp(m)*AdjMatrix1(sp(m,k),sp(m,k+1));% Calculate the weights of all the minimum hops, change * to + for the sum of the weights 
                    end
                end
                ResMatrix(i,j)=max(temp);% condition 3: choose the maximum weight among all the minimum hops
            else
                ResMatrix(i,j)=inf; % the number of the minimum hop is larger than 3
            end
        end
    end
end

ResMatrix

2、dijkstra_all.m函数文件（sp为所有的最小跳数路径集合，spcost为最小跳数）

function [sp, spcost] = dijkstra_all(AdjMatrix, s, d)
% This is an implementation of the dijkstra algorithm, wich finds the 
% minimal cost path between two nodes. It is used to solve the problem on 
% possitive weighted instances.

% the inputs of the algorithm are:
%AdjMatrix: the adjacent matrix of a graph
% s: source node index;
% d: destination node index;
n=size(AdjMatrix,1);
S(1:n) = 0;     %s, vector, set of visited vectors
dist(1:n) = inf;   % it stores the shortest distance between the source node and any other node;
prev = zeros(50,n); % Previous node, informs about the best previous node known to reach each  network node, 50 should be changed when the path is long.
count(1:n)=0;


dist(s) = 0;


while sum(S)~=n
    candidate=[];
    for i=1:n
        if S(i)==0
            candidate=[candidate dist(i)];
        else
            candidate=[candidate inf];
        end
    end
    [u_index u]=min(candidate);
    S(u)=1;
    for i=1:n
        if(dist(u)+AdjMatrix(u,u)+AdjMatrix(u,i))<dist(i)
            dist(i)=dist(u)+AdjMatrix(u,u)+AdjMatrix(u,i);
            prev(:,i)=prev(:,i).*0;
            prev(1,i)=u;
            count(i)=1;
        
        else
            if ((dist(u)+AdjMatrix(u,u)+AdjMatrix(u,i))==dist(i))&&(dist(i)~=inf)&&(u~=i)        
                if count(i)<49
                    count(i)=count(i)+1;
                end
                prev(count(i),i)=u;           
            end
        end
    end
end



sp=[];
stack=[];
num=[];
%backup
stack = [d,zeros(1,9)];
num=[1,zeros(1,9)];
spcost = dist(d);

while stack(1) ~= 0
    if stack(1)==s
        %record the path
        sp=[sp;stack];
        %pop
        stack=[stack(2:10),0];% the first element of stack is out
        num=[num(2:10),0];
        
        continue;
    end
    tmp=prev(num(1),stack(1));
    if tmp==0
        %pop
        stack=[stack(2:10),0];
        num=[num(2:10),0];
       
        continue;
    
    else
        %push
        num(1)=num(1)+1;
        stack=[tmp,stack(1:9)];
        num=[1,num(1,1:9)];
    end
   
    
end

运行main脚本文件，可得最小跳数最大权重矩阵如下：