LOJ6014「网络流 24 题 - 15」最长 k 可重区间集 坠大权不相交路径 坠大费用坠大流

大家都很强， 可与之共勉 。

题意：

  给定实直线L上n个开区间组成的集合I，和一个正整数k，试设计一个算法，从开区间集合I中选取出开区间集合 S⊆I，使得在实直线L的任何一点x， S中包含点x的开区间个数不超过k。且 ∑z∈S|z|达到最大。这样的集合S称为开区间集合I的最长 k可重区间集。∑z∈S|z|称为最长k可重区间集的长度。

  对于给定的开区间集合I和正整数k，计算开区间集合I的最长k可重区间集的长度。

题解：

  把答案看作是k条路径的权值和坠大，路径不存在边相交（此处的边就是区间）。

  既然是点可以重合，那么可以不拆点（其实拆点也可以的）。

  因为是k条路径，所以从源点到一号点连一条容量为k，费用为0的边。最后一个点向汇点，连一条容量为x(x∈[k,+∞))，的边，之后对于lr,ri之间连一条容量为1，费用为长度的边。然后我们要保证可以有区间不被选到（跳过这个区间），每个点向下一个点连一条容量为+∞（其实大于等于k就够了），然后直接跑最大费用最大流。最后一定满足条件。最后的最大流一定是k。

  画画图很好理解的

# include <bits/stdc++.h>

template < class T >  inline bool chkmax ( T& d, const T& x )  {  return d < x ? ( d = x ), 1 : 0 ;  }
template < class T >  inline bool chkmin ( T& d, const T& x )  {  return d > x ? ( d = x ), 1 : 0 ;  }

# define oo 0x3f3f3f3f

# define N 1050
# define M 5050

class MaxCostMaxFlow  {
private :
struct edge  {
int to, nxt, w, cost ;
} g [M << 1] ;

int S, T ;
int head
, dis
, pre
, ecnt ;

inline bool spfa ( int S, int T )  {
static std :: bitset < N > inq ;
static std :: deque < int > Q ;
inq.reset ( ) ; Q.clear ( ) ;
memset ( pre, 0, sizeof ( int ) * ( T + 1 ) ) ;
memset ( dis, -1, sizeof ( int ) * ( T + 1 ) ) ;
Q.push_front ( S ) ;
inq [S] = 1 ;
dis [S] = 0x3f3f3f3f ;
while ( ! Q.empty ( ) )  {
int u = Q.front ( ) ; Q.pop_front ( ) ;
inq [u] = 0 ;
for ( int i = head [u] ; i ; i = g [i].nxt )  {
int& v = g [i].to ;
if ( g [i].w && chkmax ( dis [v], dis [u] + g [i].cost ) )  {
pre [v] = i ;
if ( ! inq [v] )  {
( Q.empty ( ) || dis [v] > dis [Q.front ( )] ) ? Q.push_front ( v ) : Q.push_back ( v ) ;
inq [v] = 1 ;
}
}
}
}
return ( bool ) pre [T] ;
}
public :
MaxCostMaxFlow ( )  {  ecnt = 1 ; memset ( head, 0, sizeof head ) ;  }

inline void clear ( )  {
ecnt = 1 ; memset ( head, 0, sizeof head ) ;
}

inline void add_edge ( int u, int v, int w, int cost )  {
g [++ ecnt] = ( edge )  {  v, head [u], w, cost } ; head [u] = ecnt ;
g [++ ecnt] = ( edge )  {  u, head [v], 0, -cost } ; head [v] = ecnt ;
}

std :: pair < int, int > mcmf ( int S, int T )  {
this -> S = S, this -> T = T ;
int flow = 0, cost = 0, x ;
while ( spfa ( S, T ) )  {
x = oo ;
for ( int i = pre [T] ; i ; i = pre [g [i ^ 1].to] )  chkmin ( x, g [i].w ) ;
for ( int i = pre [T] ; i ; i = pre [g [i ^ 1].to] )  {
g [i].w -= x, g [i ^ 1].w += x ;
cost += x * g [i].cost ;
}

flow += x ;
}
return std :: make_pair ( flow, cost ) ;
}
} Lazer ;

int dust
;
int l
, r
, val
;

# undef N
# undef M

int main ( )  {
int n, k ;
scanf ( "%d%d", & n, & k ) ;
const int S = 2 * n + 1, T = 2 * n + 2 ;
for ( int i = 1 ; i <= n ; ++ i )  {
scanf ( "%d%d", l + i, r + i ) ;
if ( l [i] > r [i] )  std :: swap ( l [i], r [i] ) ;
dust [( i << 1 ) - 1] = l [i], dust [i << 1] = r [i] ;
val [i] = ( r [i] - l [i] ) ;
}
std :: sort ( dust + 1, dust + n + n + 1 ) ;
for ( int i = 1 ; i <= n ; ++ i )  {
l [i] = std :: lower_bound ( dust + 1, dust + 1 + n + n, l [i] ) - dust ;
r [i] = std :: lower_bound ( dust + 1, dust + 1 + n + n, r [i] ) - dust ;
}
Lazer.add_edge ( S, 1, k, 0 ), Lazer.add_edge ( 2 * n, T, k, 0 ) ;
for ( int i = 1 ; i <= n ; ++ i )  {
Lazer.add_edge ( l [i], r [i], 1, val [i] ) ;
}
for ( int i = 1 ; i < 2 * n ; ++ i )  {
Lazer.add_edge ( i, i + 1, oo, 0 ) ;
}
printf ( "%d\n", Lazer.mcmf ( S, T ).second ) ;
return 0 ;
}

  还有一种建图（边数复杂度是一样的）：

# include <bits/stdc++.h>

template < class T >  inline bool chkmax ( T& d, const T& x )  {  return d < x ? ( d = x ), 1 : 0 ;  }
template < class T >  inline bool chkmin ( T& d, const T& x )  {  return d > x ? ( d = x ), 1 : 0 ;  }

# define oo 0x3f3f3f3f

# define N 1050
# define M 5050

class MaxCostMaxFlow  {
private :
struct edge  {
int to, nxt, w, cost ;
} g [M << 1] ;

int S, T ;
int head
, dis
, pre
, ecnt ;

inline bool spfa ( int S, int T )  {
static std :: bitset < N > inq ;
static std :: deque < int > Q ;
inq.reset ( ) ; Q.clear ( ) ;
memset ( pre, 0, sizeof ( int ) * ( T + 1 ) ) ;
memset ( dis, -1, sizeof ( int ) * ( T + 1 ) ) ;
Q.push_front ( S ) ;
inq [S] = 1 ;
dis [S] = 0x3f3f3f3f ;
while ( ! Q.empty ( ) )  {
int u = Q.front ( ) ; Q.pop_front ( ) ;
inq [u] = 0 ;
for ( int i = head [u] ; i ; i = g [i].nxt )  {
int& v = g [i].to ;
if ( g [i].w && chkmax ( dis [v], dis [u] + g [i].cost ) )  {
pre [v] = i ;
if ( ! inq [v] )  {
( Q.empty ( ) || dis [v] > dis [Q.front ( )] ) ? Q.push_front ( v ) : Q.push_back ( v ) ;
inq [v] = 1 ;
}
}
}
}
return ( bool ) pre [T] ;
}
public :
MaxCostMaxFlow ( )  {  ecnt = 1 ; memset ( head, 0, sizeof head ) ;  }

inline void clear ( )  {
ecnt = 1 ; memset ( head, 0, sizeof head ) ;
}

inline void add_edge ( int u, int v, int w, int cost )  {
g [++ ecnt] = ( edge )  {  v, head [u], w, cost } ; head [u] = ecnt ;
g [++ ecnt] = ( edge )  {  u, head [v], 0, -cost } ; head [v] = ecnt ;
}

std :: pair < int, int > mcmf ( int S, int T )  {
this -> S = S, this -> T = T ;
int flow = 0, cost = 0, x ;
while ( spfa ( S, T ) )  {
x = oo ;
for ( int i = pre [T] ; i ; i = pre [g [i ^ 1].to] )  chkmin ( x, g [i].w ) ;
for ( int i = pre [T] ; i ; i = pre [g [i ^ 1].to] )  {
g [i].w -= x, g [i ^ 1].w += x ;
cost += x * g [i].cost ;
}

flow += x ;
}
return std :: make_pair ( flow, cost ) ;
}
} Lazer ;

int dust
;
int l
, r
, val
;

# undef N
# undef M

int main ( )  {
int n, k ;
scanf ( "%d%d", & n, & k ) ;
const int S = 2 * n + 5, T = 2 * n + 6 ;
for ( int i = 1 ; i <= n ; ++ i )  {
scanf ( "%d%d", l + i, r + i ) ;
if ( l [i] > r [i] )  std :: swap ( l [i], r [i] ) ;
dust [( i << 1 ) - 1] = l [i], dust [i << 1] = r [i] ;
val [i] = ( r [i] - l [i] ) ;
}
std :: sort ( dust + 1, dust + n + n + 1 ) ;
for ( int i = 1 ; i <= n ; ++ i )  {
l [i] = std :: lower_bound ( dust + 1, dust + 1 + n + n, l [i] ) - dust ;
r [i] = std :: lower_bound ( dust + 1, dust + 1 + n + n, r [i] ) - dust ;
}
const int S1 = 2 * n + 1, T1 = 2 * n + 2 ;
Lazer.add_edge ( S, S1, k, 0 ), Lazer.add_edge ( T1, T, k, 0 ) ;
for ( int i = 1 ; i <= 2 * n ; ++ i )  {
Lazer.add_edge ( S1, i, 1, 0 ), Lazer.add_edge ( i, T1, 1, 0 ) ;
}
for ( int i = 1 ; i <= n ; ++ i )  {
Lazer.add_edge ( l [i], r [i], 1, val [i] ) ;
}
for ( int i = 1 ; i < 2 * n ; ++ i )  {
Lazer.add_edge ( i, i + 1, oo, 0 ) ;
}
printf ( "%d\n", Lazer.mcmf ( S, T ).second ) ;
return 0 ;
}