最短路 自己覺得較快的（沒有用棧優化的）

先附上**

#include

#include
#include
#include
#include
#include
#include
#include
#define n 999999
using
namespace std;
intmain()
int u[n]
, v[n]
;double w[n]
;//放權值的平方
int m;
scanf
("%d"
,&m)
; m *=2
;for
(i =
0; i < m; i++
)// for(i = 0; i < m; i++)
// 
int s, t;
scanf
("%d %d"
,&s,
&t);
double dis[n]
;for
(i =
1; i <= n; i++
) dis[s]=0
;for
(i =
0; i < n -
1; i++)}
if(check ==0)
break;}
printf
("%.2lf\n"
, dis[t]);
return0;
}

1.第一次沒有加memset的標頭檔案

2.把公式理解錯了，以為是所有平方和後的開放（錯的）

3.理解最後的每邊鬆弛有些問題

4.n定義的太小了，因為咱也不知道有多少邊

哎，細心呀，這個演算法還是較快的，也算是好理解的，是乙個總結出的**

比鏈式前向星儲存好記一些吧！！

#include

#include
#include
#include
#include
#include
using
namespace std;
intmain()
, book[
200000]=
;int que[
200000]=
, head =
1, tail =1;
int inf =
99999999
;for
(i =
1; i <= n; i++
) dis[1]
=0;for
(i =
1; i <= n; i++
)for
(i =
1; i <= n; i++
)for
(i =
1; i <= m *
2; i++
) que[tail]=1
; tail++
; book[1]
=1;while
(head < tail)
} k = next[k];}
book[que[head]]=
0;head++;}
printf
("%d\n"
, dis[n]);
}return0;
}

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