11-matlba-bellman-ford；地杰斯特拉

%求s到各点的最短距离
function Dist = Bellman_Ford(s) 
    load cityJuli;
    for i = 1:154
        Dist(i) = inf;
    end
    Dist(s) = 0;
    flag = 1;
    for j = 1:154   %154个城市
        flag = 1;
        for i = 1 : 248 %题目给248组数据,遍历每一条边                      %用所给数据松弛
            if Dist(cityJuli(i,1)) > Dist(cityJuli(i,2)) + cityJuli(i,3)  %如果这条边的终点的dist[]值小于起始点的dist[]值加上这条边的权值，则更新dist
                Dist(cityJuli(i,1)) = Dist(cityJuli(i,2)) + cityJuli(i,3);
                flag = 0;    
            end
            if Dist(cityJuli(i,2)) > Dist(cityJuli(i,1)) + cityJuli(i,3)  %由于是无向图，故要反向松弛一遍
                Dist(cityJuli(i,2)) = Dist(cityJuli(i,1)) + cityJuli(i,3);
                flag = 0;    
            end
        end
        if(flag == 1)
            break;
        end
    end
end

%{
void bellman(int s){
    memset(d, MAX, sizeof(d));
    d[s] = 0; //必须初始化这个
    int flag = 1;
    for(int k = 0; k < n - 1; k++){
        flag = 1;
        for(int i = 0; i < m; i++){      //遍历每一条边
            if(d[u[i]] > d[v[i]] + w[i]){//如果这条边的终点的d[]值小于起始点的d[]值加上这条边的权值，则更新d
                d[u[i]] = d[v[i]] + w[i];
                flag = 0;
            } 
        }
        if(flag){
            break;
        }
    }
}
%}

function [ distance, path] = Dijk( W,st,e )  
% W  权值矩阵   st 搜索的起点   e 搜索的终点  
n=length(W);%节点数
D = W(st,:);  
for i = 1:n
    visit(i) = 1;
end
visit(st)=0;  
parent = zeros(1,n);%记录每个节点的上一个节点  
  
path =[];  
for i=1:n-1  
    temp = [];  
    %从起点出发，找最短距离的下一个点，每次不会重复原来的轨迹，设置visit判断节点是否访问  
    for j=1:n  
       if visit(j)  
           temp =[temp D(j)];  
       else  
           temp =[temp inf];  
       end  
         
    end  
      
    [value,index] = min(temp);  
     
    visit(index) = 0;  
      
    %更新 如果经过index节点，从起点到每个节点的路径长度更小，则更新，记录前趋节点，方便后面回溯循迹  
    for k=1:n  
        if D(k)>D(index)+W(index,k)  
           D(k) = D(index)+W(index,k);  
           parent(k) = index;  
        end  
    end  
      
     
end  
  
distance = D(e);%最短距离  
%回溯法  从尾部往前寻找搜索路径  
t = e;  
while t~=st && t>0  
    path =[t,path];  
    p=parent(t);t=p;  
end  
path =[st,path];%最短路径  
end