POJ2516 Minimum Cost
这题的大意是有N个店主，有M个供应商，对于N个店主，每个主有一定的需求量（对应着K种商品），M个供应商也对应着一定的供应量。然后相应的供应是有一定的费用。最后问说是否能满足需求，不满足的话输出-1，满足的话输出最小的费用。模型是比较裸，这里对于每个商品，都建个图来费用流，相应的建个超级源点和超级汇点。每次spfa去找，复杂度是O(V*E*E)。
感谢：
http://www.cppblog.com/Icyflame/archive/2009/06/30/88891.html
代码
#include
<
iostream
>
#include
<
vector
>
#include
<
queue
>
using
namespace
std;
const
int
MAX
=
55
;
const
int
INF
=
INT_MAX;
int
N, M, K;
int
need[MAX][MAX];
int
give[MAX][MAX];
int
move[MAX][MAX][MAX];
int
SS, TT;
int
mm[
2
*
MAX][
2
*
MAX];
//
flow map
int
cc[
2
*
MAX][
2
*
MAX];
//
cost map
int
prev[
2
*
MAX];
int
flow[
2
*
MAX];
int
dis[
2
*
MAX];
int
in
[
2
*
MAX];
void
ready()
{
for
(
int
i
=
1
; i
<=
N; i
++
)
for
(
int
j
=
1
; j
<=
K; j
++
) scanf(
"
%d
"
,
&
need[i][j]);
for
(
int
i
=
1
; i
<=
M; i
++
)
for
(
int
j
=
1
; j
<=
K; j
++
) scanf(
"
%d
"
,
&
give[i][j]);
for
(
int
k
=
1
; k
<=
K; k
++
)
for
(
int
i
=
1
; i
<=
N; i
++
)
for
(
int
j
=
1
; j
<=
M; j
++
) scanf(
"
%d
"
,
&
move[k][i][j]);
}
void
makeMap(
int
kind)
{
SS
=
0
, TT
=
N
+
M
+
1
;
//
make the flow map
memset(mm,
0
,
sizeof
(mm));
for
(
int
i
=
1
; i
<=
N; i
++
) mm[SS][i]
=
need[i][kind];
for
(
int
i
=
1
; i
<=
M; i
++
) mm[i
+
N][TT]
=
give[i][kind];
for
(
int
i
=
1
; i
<=
N; i
++
)
for
(
int
j
=
1
; j
<=
M; j
++
) mm[i][j
+
N]
=
need[i][kind];
//
make the cost map
memset(cc,
0
,
sizeof
(cc));
for
(
int
i
=
1
; i
<=
N; i
++
)
for
(
int
j
=
1
; j
<=
M; j
++
)
{
cc[i][j
+
N]
=
move[kind][i][j];
cc[j
+
N][i]
=
-
move[kind][i][j];
}
}
int
spfa()
{
memset(
in
,
0
,
sizeof
(
in
));
memset(dis,
-
1
,
sizeof
(dis));
memset(prev,
-
1
,
sizeof
(prev));
memset(flow,
0
,
sizeof
(flow));
in
[SS]
=
1
, dis[SS]
=
0
, flow[SS]
=
INF;
vector
<
int
>
v;
v.push_back(SS);
for
(
int
i
=
0
; i
<
v.size(); i
++
)
{
int
u
=
v[i];
in
[u]
=
0
;
for
(
int
j
=
SS; j
<=
TT; j
++
)
{
if
(mm[u][j]
>
0
&&
(dis[j]
==
-
1
||
dis[u]
+
cc[u][j]
<
dis[j]))
{
dis[j]
=
dis[u]
+
cc[u][j];
//
in[j] = 1;
prev[j]
=
u;
flow[j]
=
min(flow[u], mm[u][j]);
if
(
in
[j]
==
0
)
in
[j]
=
1
, v.push_back(j);
}
}
}
return
flow[TT];
}
int
mcmf()
{
int
res
=
0
;
while
(
1
)
{
int
now
=
spfa();
if
(now
==
0
)
break
;
int
u
=
TT;
while
(u
!=
SS)
{
int
v
=
prev[u];
mm[v][u]
-=
now;
mm[u][v]
+=
now;
res
+=
now
*
cc[v][u];
u
=
v;
}
}
for
(
int
i
=
1
; i
<=
N; i
++
)
{
if
(mm[SS][i]
>
0
)
return
-
1
;
}
return
res;
}
int
main()
{
while
(scanf(
"
%d%d%d
"
,
&
N,
&
M,
&
K), N)
{
int
res
=
0
;
ready();
for
(
int
k
=
1
; k
<=
K; k
++
)
{
makeMap(k);
int
t
=
mcmf();
if
(t
==
-
1
)
{
res
=
-
1
;
break
;
}
res
+=
t;
}
printf(
"
%d\n
"
, res);
}
}