poj 1847( floyd && spfa )

http://poj.org/problem?id=1847

一个水题，用来熟悉熟悉spfa和floyd的。

题意：有m条的铁路，要从x,到y,

之后分别就是条铁路与其他铁路的交点。第一个输入的为有n个交点。之后第一个输入的点，当前铁路到这个点是不要转向的，也就是权值为0，其余的权值都为1，求从x到y的最短路，如果到不了输出-1

裸的floyd和spfa；

 

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define inf 0x3f3f
 4 
 5 int graph[105][105];
 6 int m,n,x,y;
 7 
 8 void floyd()
 9 {
10     int k,i,j;
11     for( k = 1 ; k <= m ; k++ )
12         for( i = 1 ; i <= m ; i++ )
13             for( j = 1 ; j <= m ; j++)
14                 if( graph[ i ][ j ] > graph[ i ][ k ] + graph[ k ][ j ] )
15                     graph[ i ][ j ] = graph[ i ][ k ] + graph[ k ][ j ];
16 }
17 
18 int main()
19 {
20    // freopen("in.txt","r",stdin);
21     int tmp;
22     while(scanf("%d%d%d",&m,&x,&y)!=EOF)
23     {
24         for( int i = 1 ; i <= m ; i++ )
25             for( int j = 1 ; j <= m ; j++ )
26                 if( i == j )
27                     graph[ i ][ j ] = 0;
28                 else
29                     graph[ i ][ j ] = inf;
30         for( int i = 1 ; i <= m ; i++ )
31         {
32             scanf("%d",&n);
33             for( int j = 1 ;j <= n ; j++ )
34             {
35                 scanf("%d",&tmp);
36                 if( j == 1 )
37                     graph[ i ][ tmp ] = 0;
38                 else
39                     graph[ i ][ tmp ] = 1;
40             }
41         }
42         floyd();
43         if( graph[ x ][ y ] != inf)
44             printf("%d\n",graph[ x ][ y ]);
45         else
46             printf("-1\n");
47     }
48     return 0;
49 }
floyd
 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <queue>
 4 #define maxn 101
 5 #define inf 0x3f3f3f3f
 6 
 7 using namespace std;
 8 
 9 int m,x,n,y,pos,head[ maxn ];
10 
11 int ans , dist[ maxn ];
12 
13 bool vis[ maxn ];
14 
15 struct note {
16     int v,w,next;
17 }edge[maxn];
18 
19 void init()
20 {
21     pos = 1;
22     for(int i = 1 ;i <= n ; i++ ) dist[ i ] = inf;
23     memset( head , -1 , sizeof( head ) );
24     memset( vis , false ,sizeof( vis ) );
25 }
26 
27 void add(int x,int v,int w)
28 {
29     edge[ pos ].v = v;
30     edge[ pos ].w = w;
31     edge[ pos ].next = head[ x ];
32     head[ x ] = pos++;
33 }
34 
35 void spfa()
36 {
37     queue<int >s;
38     s.push(x);
39     vis[ x ] = true;
40     dist[ x ] = 0;
41     while(!s.empty())
42     {
43         int tmp = s.front();
44         s.pop();
45         vis [ tmp ] = false;
46         for( int i = head[ tmp ] ; i != -1 ; i = edge[ i ].next )
47         {
48             if( dist[ edge[ i ].v ] > dist[ tmp ] + edge[ i ].w)
49             {
50                 dist[ edge[ i ].v ] = dist[ tmp ] + edge[ i ].w;
51                 if( !vis[ edge[ i ].v ] )
52                 {
53                     s.push( edge[ i ].v );
54                     vis[ edge[ i ].v ] =true;
55                 }
56             }
57 
58         }
59     }
60 }
61 
62 int main()
63 {
64   //  freopen("in.txt","r",stdin);
65     while(~scanf("%d%d%d",&n,&x,&y))
66     {
67         init();
68         int tmp;
69         for(int j = 1 ; j <= n ; j++ )
70         {
71             scanf("%d",&m);
72             for(int i = 1 ; i <= m ; i++ )
73             {
74                 scanf("%d",&tmp);
75                 if( i == 1 ) add( j , tmp , 0);
76                 else add( j , tmp , 1 );
77             }
78         }
79         spfa();
80         if( dist [ y ] == inf ) printf("-1\n");
81         else printf("%d\n",dist[ y ]);
82     }
83     return 0;
84 }
spfa

 

posted @ 2016-08-05 11:02  一个_小菜鸟  阅读(164)  评论(0)  编辑  收藏  举报