学步园

传送门：【HDU】3631 Shortest Path

题目分析：考察对floyd的理解，如果存在一条路i->k,k->j，那么如果中间的k是未被标记的，那么我们就不用k去更新i->j之间的最短路，那么除非存在另一条路i->l,l->j且l已经被标记或者存在i->j的边，否则i->j是不可达的。每次标记一个点的时候就用这个点去更新所有的最短路即可。不要忘记重边的判断。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define CLR( a , x ) memset ( a , x , sizeof a )

typedef long long LL ;

const int MAXN = 300 ;
const int INF = 0x3f3f3f3f ;

bool vis[MAXN] ;
int G[MAXN][MAXN] ;
int n , m , q ;

void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}

void solve () {
	int u , v , c , tmp ;
	CLR ( G , INF ) ;
	CLR ( vis , 0 ) ;
	while ( m -- ) {
		scanf ( u ) , scanf ( v ) , scanf ( c ) ;
		if ( G[u][v] > c ) G[u][v] = c ;
	}
	while ( q -- ) {
		scanf ( c ) ;
		if ( c == 0 ) {
			scanf ( u ) ;
			if ( vis[u] ) printf ( "ERROR! At point %d\n" , u ) ;
			else {
				vis[u] = 1 ;
				G[u][u] = 0 ;
				REP ( i , 0 , n ) REP ( j , 0 , n )
					if ( G[i][j] > G[i][u] + G[u][j] ) G[i][j] = G[i][u] + G[u][j] ;
			}
		} else {
			scanf ( u ) , scanf ( v ) ;
			if ( !vis[u] || !vis[v] ) printf ( "ERROR! At path %d to %d\n" , u , v ) ;
			else if ( G[u][v] == INF ) printf ( "No such path\n" ) ;
			else printf ( "%d\n" , G[u][v] ) ;
		}
	}		
}

int main () {
	int cas = 0 ;
	while ( ~scanf ( "%d%d%d" , &n , &m , &q ) && ( n || m || q ) ) {
		if ( cas ) printf ( "\n" ) ;
		printf ( "Case %d:\n" , ++ cas ) ;
		solve () ;
	}
	return 0 ;
}