118 lines
2.9 KiB
C++
118 lines
2.9 KiB
C++
#include <cstdio>
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#include <cstring>
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#include <algorithm>
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using namespace std ;
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#define REP( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
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#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
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#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
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#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
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#define CLR( a , x ) memset ( a , x , sizeof a )
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typedef long long LL ;
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const int MAXN = 55 ;
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const int MAXE = 205 ;
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const double INF = 1e10 ;
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const double eps = 1e-8 ;
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struct Edge {
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int v ;
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double c ;
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Edge* next ;
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} E[MAXE] , *H[MAXN] , *cur ;
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struct Line {
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int x , y , a , b ;
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double v ;
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void input () {
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scanf ( "%d%d%d%d" , &x , &y , &a , &b ) ;
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}
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bool operator < ( const Line& p ) const {
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return v > p.v ;
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}
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} L[MAXE] ;
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int n , m , t ;
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int p[MAXN] ;
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double f[MAXN] ;
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double a[10005] ;
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int cnt ;
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double st , ed , ans ;
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int sgn ( double x ) {
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return ( x > eps ) - ( x < -eps ) ;
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}
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void clear () {
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cur = E ;
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CLR ( H , 0 ) ;
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FOR ( i , 1 , n ) {
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p[i] = i ;
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f[i] = INF ;
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}
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}
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void addedge ( int u , int v , double c ) {
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cur -> v = v ;
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cur -> c = c ;
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cur -> next = H[u] ;
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H[u] = cur ++ ;
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}
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int find ( int x ) {
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return p[x] == x ? x : ( p[x] = find ( p[x] ) ) ;
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}
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void kruskal () {
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sort ( L , L + m ) ;
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int cnt = 1 ;
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REP ( i , 0 , m ) {
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int x = find ( L[i].x ) ;
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int y = find ( L[i].y ) ;
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if ( x != y ) {
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p[x] = y ;
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addedge ( L[i].x , L[i].y , L[i].v ) ;
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addedge ( L[i].y , L[i].x , L[i].v ) ;
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if ( ++ cnt == n ) break ;
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}
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}
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}
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void dfs ( int u , int fa ) {
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travel ( e , H , u ) {
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int v = e -> v ;
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if ( v == fa ) continue ;
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f[v] = min ( f[u] , e -> c ) ;
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dfs ( v , u ) ;
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}
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}
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int unique ( int n ) {
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int cnt = 1 ;
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sort ( a + 1 , a + n + 1 ) ;
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FOR ( i , 2 , n ) if ( sgn ( a[i] - a[cnt] ) ) a[++ cnt] = a[i] ;
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return cnt ;
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}
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void solve () {
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ans = 0 ;
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cnt = 0 ;
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scanf ( "%d%d%d" , &n , &m , &t ) ;
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REP ( i , 0 , m ) {
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L[i].input () ;
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REP ( j , 0 , i ) {
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if ( L[i].b == L[j].b ) continue ;
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a[++ cnt] = ( double ) ( L[i].a - L[j].a ) / ( L[j].b - L[i].b ) ;
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}
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}
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a[++ cnt] = t ;
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cnt = unique ( cnt ) ;
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st = 0 ;
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FOR ( i , 1 , cnt ) {
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if ( a[i] < 0 ) continue ;
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if ( a[i] > t ) break ;
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ed = a[i] ;
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double mid = ( st + ed ) / 2 ;
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REP ( j , 0 , m ) L[j].v = L[j].a + L[j].b * mid ;
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clear () ;
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kruskal () ;
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dfs ( 1 , 0 ) ;
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FOR ( j , 2 , n ) ans += f[j] * ( ed - st ) ;
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st = ed ;
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}
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printf ( "%.3f\n" , ans / t ) ;
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}
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int main () {
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int T ;
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scanf ( "%d" , &T ) ;
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while ( T -- ) solve () ;
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return 0 ;
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}
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