題目鏈接
題意
NN N 這條路兩端的人口總數修路的花費\frac{這條路兩端的人口總數}{修路的花費} 修 路 的 花 費 這 條 路 兩 端 的 人 口 總 數 ?
思路
先求最小生成樹然后判斷每條邊，如果這條邊是最小生成樹上的邊最小生數的權值和?這條邊這條邊兩端人口總和\frac{最小生數的權值和-這條邊}{這條邊兩端人口總和} 這 條 邊 兩 端 人 口 總 和 最 小 生 數 的 權 值 和 ? 這 條 邊 ? MaxMax M a x
kruskal：kruskal： k r u s k a l ： 218MS 32720K ，空間勉強卡過，vectorvector v e c t o r prim：prim： p r i m ： 124MS 10268K
#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
#define endl '\n'
const int maxn
= 1001 ;
const int inf
= 0x3f3f3f3f ;
const int mod
= 1e9 + 7 ;
using namespace std
;
int n
, m
;
struct ac
{ int u
, v
, w
; ac ( int u_
= 0 , int v_
= 0 , int w_
= 0 ) { u
= u_
; v
= v_
; w
= w_
; } inline int Dis ( ac x
) { return ( ( x
. u
- u
) * ( x
. u
- u
) + ( x
. v
- v
) * ( x
. v
- v
) ) ; }
} ;
int g
[ maxn
] [ maxn
] , val
[ maxn
] , vis
[ maxn
] , dis
[ maxn
] ;
int pre
[ maxn
] , maxd
[ maxn
] [ maxn
] ;
bool used
[ maxn
] [ maxn
] ; double Count ( int i
, int j
, double sum
) { return ( val
[ i
] + val
[ j
] ) * 1.0 / ( sum
- sqrt ( g
[ i
] [ j
] ) ) ;
}
void prim ( int s
) { mem ( maxd
, 0 ) ; mem ( vis
, 0 ) ; mem ( used
, 0 ) ; for ( int i
= 0 ; i
< n
; ++ i
) { dis
[ i
] = g
[ s
] [ i
] ; pre
[ i
] = s
; } vis
[ s
] = 1 ; double sum
= 0 ; for ( int i
= 1 ; i
< n
; ++ i
) { int u
= - 1 , MIN
= inf
; for ( int j
= 0 ; j
< n
; ++ j
) { if ( vis
[ j
] ) continue ; if ( MIN
> dis
[ j
] ) { MIN
= dis
[ j
] ; u
= j
; } } if ( u
== - 1 ) break ; vis
[ u
] = 1 ; sum
+ = sqrt ( MIN
) ; used
[ pre
[ u
] ] [ u
] = used
[ u
] [ pre
[ u
] ] = 1 ; maxd
[ u
] [ pre
[ u
] ] = maxd
[ pre
[ u
] ] [ u
] = MIN
; for ( int j
= 0 ; j
< n
; ++ j
) { if ( j
== u
) continue ; if ( vis
[ j
] ) { maxd
[ u
] [ j
] = maxd
[ j
] [ u
] = max ( maxd
[ pre
[ u
] ] [ j
] , MIN
) ; } if ( vis
[ j
] == 0 && dis
[ j
] > g
[ u
] [ j
] ) { dis
[ j
] = g
[ u
] [ j
] ; pre
[ j
] = u
; } } } double ans
= 0 ; for ( int i
= 0 ; i
< n
; ++ i
) { for ( int j
= i
+ 1 ; j
< n
; ++ j
) { if ( used
[ i
] [ j
] ) ans
= max ( ans
, Count ( i
, j
, sum
) ) ; else ans
= max ( ans
, Count ( i
, j
, sum
- sqrt ( maxd
[ i
] [ j
] ) + sqrt ( g
[ i
] [ j
] ) ) ) ; } } printf ( "%.2lf\n" , ans
) ;
} int main ( ) { ios
:: sync_with_stdio ( false ) ; cin
. tie ( 0 ) ; cout
. tie ( 0 ) ; int T
; scanf ( "%d" , & T
) ; while ( T
-- ) { scanf ( "%d" , & n
) ; vector
< ac
> t ( n
) ; mem ( g
, inf
) ; for ( int i
= 0 ; i
< n
; ++ i
) { scanf ( "%d%d%d" , & t
[ i
] . u
, & t
[ i
] . v
, & val
[ i
] ) ; } for ( int i
= 0 ; i
< n
; ++ i
) { for ( int j
= i
+ 1 ; j
< n
; ++ j
) { g
[ i
] [ j
] = g
[ j
] [ i
] = t
[ i
] . Dis ( t
[ j
] ) ; } } prim ( 0 ) ; } return 0 ;
}
#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#define lowbit(x) (x & -x)
#define mem(a, b) memset(a, b, sizeof(a))
#define mid ((l + r) >> 1)
#define lc rt<<1
#define rc rt<<1|1
#define endl '\n'
const int maxn
= 1e3 + 1 ;
const int inf
= 0x3f3f3f3f ;
const int mod
= 1e9 + 7 ;
using namespace std
;
struct ac
{ int u
, v
, flag
; double w
; ac ( int u_
= 0 , int v_
= 0 , double w_
= 0 , int flag_
= 0 ) { u
= u_
; v
= v_
; w
= w_
; flag
= flag_
; } bool operator < ( ac t
) { return w
< t
. w
; }
} g
[ 1000001 ] ;
vector
< int > son
[ maxn
] ;
int pre
[ maxn
] , val
[ maxn
] ;
double dis
[ maxn
] [ maxn
] ;
int find
( int x
) { return ( pre
[ x
] == x
) ? x
: pre
[ x
] = find ( pre
[ x
] ) ;
}
int n
, m
;
double Dis ( ac x
, ac y
) { return sqrt ( ( x
. u
- y
. u
) * ( x
. u
- y
. u
) + ( x
. v
- y
. v
) * ( x
. v
- y
. v
) ) ;
}
double Count ( double all
, ac t
) { return ( val
[ t
. u
] + val
[ t
. v
] ) * 1.0 / ( all
- t
. w
) ;
}
void Kruskal ( ) { for ( int i
= 0 ; i
<= n
; ++ i
) { son
[ i
] . clear ( ) ; son
[ i
] . push_back ( i
) ; pre
[ i
] = i
; } sort ( g
, g
+ m
) ; double sum
= 0 ; int cnt
= 0 ; for ( int i
= 0 ; i
< m
; ++ i
) { if ( cnt
== n
+ 1 ) break ; int fx
= find ( g
[ i
] . u
) ; int fy
= find ( g
[ i
] . v
) ; if ( fx
== fy
) continue ; g
[ i
] . flag
= 1 ; sum
+ = g
[ i
] . w
; cnt
++ ; int lenx
= son
[ fx
] . size ( ) ; int leny
= son
[ fy
] . size ( ) ; if ( lenx
< leny
) { swap ( lenx
, leny
) ; swap ( fx
, fy
) ; } for ( int j
= 0 ; j
< lenx
; ++ j
) { for ( int k
= 0 ; k
< leny
; ++ k
) { dis
[ son
[ fx
] [ j
] ] [ son
[ fy
] [ k
] ] = dis
[ son
[ fy
] [ k
] ] [ son
[ fx
] [ j
] ] = g
[ i
] . w
; } } pre
[ fy
] = fx
; for ( int j
= 0 ; j
< leny
; ++ j
) { son
[ fx
] . push_back ( son
[ fy
] [ j
] ) ; } son
[ fy
] . clear ( ) ; } double ans
= - 1 ; for ( int i
= 0 ; i
< m
; ++ i
) { if ( g
[ i
] . flag
) { ans
= max ( ans
, Count ( sum
, g
[ i
] ) ) ; } else { ans
= max ( ans
, Count ( sum
+ g
[ i
] . w
- dis
[ g
[ i
] . u
] [ g
[ i
] . v
] , g
[ i
] ) ) ; } } printf ( "%.2lf\n" , ans
) ;
} int main ( ) { ios
:: sync_with_stdio ( false ) ; cin
. tie ( 0 ) ; cout
. tie ( 0 ) ; int T
; scanf ( "%d" , & T
) ; while ( T
-- ) { scanf ( "%d" , & n
) ; vector
< ac
> t ( n
) ; for ( int i
= 0 ; i
< n
; ++ i
) { scanf ( "%d%d%d" , & t
[ i
] . u
, & t
[ i
] . v
, & val
[ i
] ) ; } m
= 0 ; for ( int i
= 0 ; i
< n
; ++ i
) { for ( int j
= i
+ 1 ; j
< n
; ++ j
) { g
[ m
++ ] = ac ( i
, j
, Dis ( t
[ i
] , t
[ j
] ) , 0 ) ; } } Kruskal ( ) ; } return 0 ;
}
總結
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