【POJ】2472 106 miles to Chicago 最短路

传送门：【POJ】2472 106 miles to Chicago

题目分析：直接将边的权值c变成log(c/100.0)，这样将乘积最大转化为相加最大了，这样求的就是最长路了，注意log里面一定要除以100，否则会产生负权环。最后输出exp(ans)*100即可。

代码如下：

#include <cmath>
#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 double type_c ;

const int MAXN = 105 ;
const int MAXH = 20005 ;
const int MAXE = 20005 ;
const double INF = 1e9 ;

struct Edge {
	int v ;
	type_c c ;
	Edge* next ;
} ;

struct Shortest_Path_Algorithm {
	Edge E[MAXE] , *H[MAXN] , *cur ;
	type_c d[MAXN] ;
	bool vis[MAXN] ;
	int Q[MAXN] , head , tail ;
	void init () {
		cur = E ;
		CLR ( H , 0 ) ;
	}
	void addedge ( int u , int v , type_c c ) {
		cur -> v = v ;
		cur -> c = c ;
		cur -> next = H[u] ;
		H[u] = cur ++ ;
	}
	void spfa ( int s ) {
		head = tail = 0 ;
		REP ( i , 0 , MAXN ) d[i] = INF ;
		CLR ( vis , 0 ) ;
		d[s] = 0 ;
		Q[tail ++] = s ;
		while ( head != tail ) {
			int u = Q[head ++] ;
			if ( head == MAXN ) head = 0 ;
			vis[u] = 0 ;
			travel ( e , H , u ) {
				int v = e -> v ;
				if ( d[v] > d[u] + e -> c ) {
					d[v] = d[u] + e -> c ;
					if ( !vis[v] ) {
						vis[v] = 1 ;
						if ( d[v] < d[Q[head]] ) {
							if ( head == 0 ) head = MAXN ;
							Q[-- head] = v ;
						} else {
							Q[tail ++] = v ;
							if ( tail == MAXN ) tail = 0 ;
						}
					}
				}
			}
		}
	}
} G ;

int n , m ;

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 ;
	G.init () ;
	while ( m -- ) {
		scanf ( "%d%d%d" , &u , &v , &c ) ;
		G.addedge ( u , v , -log ( c / 100.0 ) ) ;
		G.addedge ( v , u , -log ( c / 100.0 ) ) ;
	}
	G.spfa ( 1 ) ;
	printf ( "%.6f percent\n" , exp ( -G.d
 ) * 100 ) ;
}

int main () {
	while ( ~scanf ( "%d%d" , &n , &m ) && n ) solve () ;
	return 0 ;
}