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；

#include <stdio.h>
#include <string.h>
#define inf 0x3f3f

int graph[105][105];
int m,n,x,y;

void floyd()
{
    int k,i,j;
    for( k = 1 ; k <= m ; k++ )
        for( i = 1 ; i <= m ; i++ )
            for( j = 1 ; j <= m ; j++)
                if( graph[ i ][ j ] > graph[ i ][ k ] + graph[ k ][ j ] )
                    graph[ i ][ j ] = graph[ i ][ k ] + graph[ k ][ j ];
}

int main()
{
   // freopen("in.txt","r",stdin);
    int tmp;
    while(scanf("%d%d%d",&m,&x,&y)!=EOF)
    {
        for( int i = 1 ; i <= m ; i++ )
            for( int j = 1 ; j <= m ; j++ )
                if( i == j )
                    graph[ i ][ j ] = 0;
                else
                    graph[ i ][ j ] = inf;
        for( int i = 1 ; i <= m ; i++ )
        {
            scanf("%d",&n);
            for( int j = 1 ;j <= n ; j++ )
            {
                scanf("%d",&tmp);
                if( j == 1 )
                    graph[ i ][ tmp ] = 0;
                else
                    graph[ i ][ tmp ] = 1;
            }
        }
        floyd();
        if( graph[ x ][ y ] != inf)
            printf("%d
",graph[ x ][ y ]);
        else
            printf("-1
");
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
#include <queue>
#define maxn 101
#define inf 0x3f3f3f3f

using namespace std;

int m,x,n,y,pos,head[ maxn ];

int ans , dist[ maxn ];

bool vis[ maxn ];

struct note {
    int v,w,next;
}edge[maxn];

void init()
{
    pos = 1;
    for(int i = 1 ;i <= n ; i++ ) dist[ i ] = inf;
    memset( head , -1 , sizeof( head ) );
    memset( vis , false ,sizeof( vis ) );
}

void add(int x,int v,int w)
{
    edge[ pos ].v = v;
    edge[ pos ].w = w;
    edge[ pos ].next = head[ x ];
    head[ x ] = pos++;
}

void spfa()
{
    queue<int >s;
    s.push(x);
    vis[ x ] = true;
    dist[ x ] = 0;
    while(!s.empty())
    {
        int tmp = s.front();
        s.pop();
        vis [ tmp ] = false;
        for( int i = head[ tmp ] ; i != -1 ; i = edge[ i ].next )
        {
            if( dist[ edge[ i ].v ] > dist[ tmp ] + edge[ i ].w)
            {
                dist[ edge[ i ].v ] = dist[ tmp ] + edge[ i ].w;
                if( !vis[ edge[ i ].v ] )
                {
                    s.push( edge[ i ].v );
                    vis[ edge[ i ].v ] =true;
                }
            }

        }
    }
}

int main()
{
  //  freopen("in.txt","r",stdin);
    while(~scanf("%d%d%d",&n,&x,&y))
    {
        init();
        int tmp;
        for(int j = 1 ; j <= n ; j++ )
        {
            scanf("%d",&m);
            for(int i = 1 ; i <= m ; i++ )
            {
                scanf("%d",&tmp);
                if( i == 1 ) add( j , tmp , 0);
                else add( j , tmp , 1 );
            }
        }
        spfa();
        if( dist [ y ] == inf ) printf("-1
");
        else printf("%d
",dist[ y ]);
    }
    return 0;
}