%求最短路徑的Floyd算法實現
%1先區別有向圖和無向圖，2輸入頂點數和邊數并檢查合法性，3輸入每邊的起點、終點、權重并檢查合法性，并初始化鄰接矩陣和路徑矩陣，4調用自定義函數Floyd

%1
bool=0;
while?bool<1
????type=input（‘輸入圖的種類1代表無向圖2代表有向圖>>‘）;
????if?type==1?||?type==2
????????bool=1;
????else
????????display（‘請重新輸入圖像類型‘）;
????end
end

%2
boolean=0;
while?boolean<1
????np=input（‘輸入頂點個數>>‘）;
????nl=input（‘輸入路徑數>>‘）;
????if?nl<=type*np*（np-1）/2;
????????boolean=1;
????else
????????display（‘輸入的頂點數和邊數不合法，請重新輸入‘）;
????end
end

%3
inf=9999;
D=inf*ones（npnp）;
for?i=1:np
????D（ii）=0;
end
P=ones（npnp）;
for?i=1:np
????P（:i）=i;
end
i=1;
while?i<=nl
????a=input（‘以[sew]形式輸入每邊的起點編號（從1開始）、終點編號、權重‘）;
????startp=a（11）;
????endp=a（12）;
????weight=a（13）;
????if?startp<=0?||?endp<=0?||?weight<=0?||?startp>np?||endp>np
????????display（‘本組數據不合法，請重新輸入本組數據‘）;
????else?if?type==1
????????D（startpendp）=weight;
????????D（endpstartp）=weight;??????????????????????????????????????????%可以證明無向圖的鄰接矩陣是對稱矩陣
????????end
????????if?type==2
????????????D（startpendp）=weight;
????????end
????????i=i+1;
????end
end

%4
if?type==1
????Floyd1（npDP）;
else
????Floyd2（npDP）;
end

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件???????1334??2018-08-29?14:17??floyd.m

?????文件???????1227??2018-08-29?14:33??Floyd1.m

?????文件????????955??2018-08-29?14:37??Floyd2.m

-----------?---------??----------?-----??----

?????????????????3516????????????????????3