| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9851 | Accepted: 3375 |
Description
Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time.
Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (conveniently numbered 1.. L) and the P (2 ≤ P ≤ 5000) unidirectional cow paths that join them. Farmer John will drive the cows to a starting landmark of their choice, from which they will walk along the cow paths to a series of other landmarks, ending back at their starting landmark where Farmer John will pick them up and take them back to the farm. Because space in the city is at a premium, the cow paths are very narrow and so travel along each cow path is only allowed in one fixed direction.
While the cows may spend as much time as they like in the city, they do tend to get bored easily. Visiting each new landmark is fun, but walking between them takes time. The cows know the exact fun values Fi (1 ≤ Fi ≤ 1000) for each landmark i.
The cows also know about the cowpaths. Cowpath i connects landmark L1i to L2i (in the direction L1i -> L2i ) and requires time Ti (1 ≤ Ti ≤ 1000) to traverse.
In order to have the best possible day off, the cows want to maximize the average fun value per unit time of their trip. Of course, the landmarks are only fun the first time they are visited; the cows may pass through the landmark more than once, but they do not perceive its fun value again. Furthermore, Farmer John is making the cows visit at least two landmarks, so that they get some exercise during their day off.
Help the cows find the maximum fun value per unit time that they can achieve.
Input
* Line 1: Two space-separated integers: L and P
* Lines 2..L+1: Line i+1 contains a single one integer: Fi
* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L1i , L2i , and Ti
Output
* Line 1: A single number given to two decimal places (do not perform explicit rounding), the maximum possible average fun per unit time, or 0 if the cows cannot plan any trip at all in accordance with the above rules.
Sample Input
5 7
30
10
10
5
10
1 2 3
2 3 2
3 4 5
3 5 2
4 5 5
5 1 3
5 2 2
Sample Output
6.00
Source
USACO 2007 December Gold也不是很难的题，套路一下 ，唯一就是求是否存在正环的时候，可以把边权取反，用spfa搞 。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
const int N = + , M = + ;
#define inf 1000000000
using namespace std ;
int cnt , n , m , node[N] , head[N] , cs[N] ; double r ;
bool used[N] ; double dis[N] ;
struct id
{
int fro , to , nxt , val ; double w ;
} edge[M] ;
queue< int > Q ; void add( int u , int v , int c )
{
edge[++cnt].fro = u , edge[cnt].to = v ;
edge[cnt].val = c , edge[cnt].nxt = head[u] ; head[u] = cnt ;
} void Init( )
{
scanf( "%d%d" , &n , &m ) ;
for( int x = ; x <= n ; ++x )
{
scanf( "%d" , node + x ) ;
r += node[x] ;
}
int u , v , c ;
for( int x = ; x <= m ; ++x )
{
scanf( "%d%d%d" , &u , &v , &c ) ;
add( u , v , c ) ;
}
} bool spfa( double l )
{
while( !Q.empty( ) ) Q.pop( ) ;
for( int x = ; x <= n ; ++x ) dis[x] = inf ;
for( int x = ; x <= cnt ; ++x ) edge[x].w = edge[x].val * l - node[edge[x].fro] ;
memset( cs , , sizeof(cs) ) ;
memset( used , , sizeof( used ) ) ;
Q.push( ) ; dis[] = ; used[] = true ;
while( !Q.empty( ) )
{
int u = Q.front( ) ; Q.pop( ) ;used[u] = false ; ++cs[u] ;
if( cs[u] == n ) return ;
for( int i = head[u] ; i ; i = edge[i].nxt )
{
int v = edge[i].to ;
if( dis[v] >= dis[u] + edge[i].w )
{
dis[v] = dis[u] + edge[i].w ;
if( !used[v] ) used[v] = true , Q.push( v ) ;
} }
}
return false ;
} void Sovle( )
{
double l = ;
while( r - l > 1e- )
{
double mid = l + ( r - l ) / ;
if( spfa( mid ) ) l = mid ;
else r = mid ;
}
printf( "%.2f\n" , l ) ;
} int main( )
{
Init( ) ;
Sovle( ) ;
return ;
}
