CFR 653 D. Delivery Bears ( Flow, Binary Search )

Problem - 653D - Codeforces

N 個節點 M 條有向邊的圖。有 X 隻熊。起點 1，終點 N。

2 ≤ N ≤ 50
1 ≤ M ≤ 500
1 ≤ X ≤ 1e5
1 ≤ C[ i ] ≤ 1e6

O( O( Dinic's ) * 100 )

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#pragma GCC optimize ("fast-math")

#include <bits/stdc++.h>
using namespace std;

template< class T >
struct Dinic {
static const int MAXV = 50;
static const T INF = 0x3f3f3f3f;
struct Edge {
int v;
T f;
int re;
Edge( int _v, T _f, int _re ): v( _v ), f( _f ), re( _re ) {}
};
int n, s, t, level[ MAXV ];
vector< Edge > E[ MAXV ];
int now[ MAXV ];
Dinic( int _n, int _s, int _t ): n( _n ), s( _s ), t( _t ) {}
void add_edge( int u, int v, T f, bool bidirectional = false ) {
E[ u ].emplace_back( v, f, E[ v ].size() );
E[ v ].emplace_back( u, 0, E[ u ].size() - 1 );
if( bidirectional ) {
E[ v ].emplace_back( u, f, E[ u ].size() - 1 );
}
}
bool BFS() {
memset( level, -1, sizeof( level ) );
queue< int > que;
que.emplace( s );
level[ s ] = 0;
while( not que.empty() ) {
int u = que.front();
que.pop();
for( auto it: E[ u ] ) {
if( it.f > 0 and level[ it.v ] == -1 ) {
level[ it.v ] = level[ u ] + 1;
que.emplace( it.v );
}
}
}
return level[ t ] != -1;
}
T DFS( int u, T nf ) {
if( u == t ) return nf;
T res = 0;
while( now[ u ] < E[ u ].size() ) {
Edge &it = E[ u ][ now[ u ] ];
if( it.f > 0 and level[ it.v ] == level[ u ] + 1 ) {
T tf = DFS( it.v, min( nf, it.f ) );
res += tf; nf -= tf; it.f -= tf;
E[ it.v ][ it.re ].f += tf;
if( nf == 0 ) return res;
} else ++now[ u ];
}
if( not res ) level[ u ] = -1;
return res;
}
T flow( T res = 0 ) {
while( BFS() ) {
T temp;
memset( now, 0, sizeof( now ) );
while( temp = DFS( s, INF ) ) {
res += temp;
res = min( res, INF );
}
}
return res;
}
};

const int MAXN = 50;
const int MAXM = 500;

int N, M, X;
int A[ MAXM ], B[ MAXM ], C[ MAXM ];

signed main() {
ios::sync_with_stdio( 0 );
cin >> N >> M >> X;
for( int i = 0; i < M; ++i ) {
cin >> A[ i ] >> B[ i ] >> C[ i ];
--A[ i ], --B[ i ];
}
double lb = 0.0, ub = 1e6;
for( int i = 0; i < 100; ++i ) {
double mid = ( lb + ub ) / 2;
Dinic< int > din( N, 0, N - 1 );
for( int i = 0; i < M; ++i ) {
din.add_edge( A[ i ], B[ i ], min( C[ i ] / mid, 1e9 ) );
}
( din.flow() >= X ? lb : ub ) = mid;
}
cout << fixed << setprecision( 7 ) << lb * X << endl;
return 0;
}