hdu1595-find the longest of the shortest

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http://acm.hdu.edu.cn/showproblem.php?pid=1595

求最短路中间的最长路径

题意为求最短路，但是要求的是最短路中间当去掉某一条路时的情况下路径最长的，这里是使用Dijkstra写的，代码相较其他简便了很多，值得注意的是当记录路径时，当传入的v0为0 时不要记录

#include<iostream>#include<cstdio>#include<cstring>const int INF = 1000000000 ;const int maxn = 1005 ;int edge[ maxn ][ maxn ] ;int path[ maxn ] ;int used[ maxn ] ;int dist[ maxn ] ;int n , m ;int Dijkstra( int v0 ){int  i , j , k ;memset( used , 0 , sizeof( used ) ) ;for( i = 0 ; i <= n ; i++ ) {dist[ i ] = INF ;}dist[ 1 ] = 0 ;for( i = 1 ; i <= n ; i++ ){int MIN = INF , u = v0 ;for( j = 1 ; j <= n ; j++ ){if( !used[ j ] && dist[ j ] < MIN ){MIN = dist[ j ] ;u = j ;}}used[ u ] = 1 ;for( k = 1 ; k <= n ; k++ ){if( !used[ k ] && edge[ u ][ k ] && edge[ u ][ k ] + dist[ u ] < dist[ k ] ){dist[ k ] = dist[ u ] + edge[ u ][ k ] ;if( v0 )path[ k ] = u ;}}}}int main(){int i , j , k ;int x , y , temp ;while( scanf( "%d%d" , &n ,&m ) != EOF ){memset( edge , 0 , sizeof( edge ) ) ;for( i = 1 ; i <= n ; i++ ){for( j = 1 ; j <= n ; j++ ){if( i == j )edge[ i ][ j ] = 0 ;elseedge[ i ][ j ] = INF ;}}memset( path , 0 , sizeof( path ) ) ;for( i = 0 ; i < m ; i++ ){scanf( "%d%d%d" , &x, &y , &temp ) ;if( temp < edge[ x ][ y ] )edge[ x ][ y ] = edge[ y ][ x ] = temp ;}Dijkstra( 1 ) ;int MAX = dist[ n ] ;for( i = n ; i != 1 ; i = path[ i ] ){int t = edge[ path[ i ] ][ i ] ;edge[ path[ i ] ][ i ] = INF ;edge[ i ][ path[ i ] ] = INF ;Dijkstra( 0 ) ; if( dist[ n ] > MAX )MAX = dist[ n ] ;edge[ i ][ path[ i ] ] = edge[ path[ i ] ][ i ] = t ;}printf( "%d\n" , MAX ) ;}return 0 ;}