POJ 1459 Power Network
解题思路：
首先添加两个节点，分别为起始点S,终点T，如果节点为power station，则添加一条S到当前节点的路径(路径容量为该节点的生产能力)；如果节点为consumer，则添加当前节点到T的一条路径(路径容量为该节点的消费能力)，则问题转化为求解S到T的最大流问题。
代码如下:
#include
<
iostream
>
using
namespace
std;
#define
INF 1e8
#define
MAX_VECT 205
#define
MAX_EDGE 22000
int
to[MAX_EDGE], next[MAX_EDGE], cap[MAX_EDGE], m;
int
v[MAX_VECT], d[MAX_VECT], queue[MAX_VECT], n;
int
S, T;
inline
void
single_insert(
int
_u,
int
_v,
int
var)
{
to[m]
=
_v;
cap[m]
=
var;
next[m]
=
v[_u];
v[_u]
=
m
++
;
}
void
insert(
int
from,
int
to,
int
cap)
{
single_insert(from, to, cap);
single_insert(to, from,
0
);
}
bool
bfs_initial()
{
memset(d,
-
1
,
sizeof
(d));
int
bg, ed, x, y;
bg
=
ed
=
d[S]
=
0
;
queue[ed
++
]
=
S;
while
(bg
<
ed)
{
x
=
queue[bg
++
];
for
(
int
i
=
v[x]; i
+
1
; i
=
next[i])
{
y
=
to[i];
if
(cap[i]
&&
d[y]
==
-
1
)
{
d[y]
=
d[x]
+
1
;
if
(y
==
T)
return
true
;
queue[ed
++
]
=
y;
}
}
}
return
false
;
}
int
dinic()
{
int
ans
=
0
;
while
(bfs_initial())
{
int
edge, x, y, back, iter
=
1
;
while
(iter)
{
x
=
(iter
==
1
)
?
S : to[queue[iter
-
1
]];
if
(x
==
T)
{
int
minE, minCap
=
INF;
for
(
int
i
=
1
; i
<
iter; i
++
)
{
edge
=
queue[i];
if
(cap[edge]
<
minCap)
{
minCap
=
cap[edge];
back
=
i;
}
}
for
(
int
i
=
1
; i
<
iter; i
++
)
{
edge
=
queue[i];
cap[edge]
-=
minCap;
cap[edge
^
1
]
+=
minCap;
}
ans
+=
minCap;
iter
=
back;
}
else
{
for
(edge
=
v[x]; edge
+
1
; edge
=
next[edge])
{
y
=
to[edge];
if
(cap[edge]
&&
d[y]
==
d[x]
+
1
)
break
;
}
if
(edge
+
1
)
queue[iter
++
]
=
edge;
else
{
d[x]
=
-
1
;
iter
--
;
}
}
}
}
return
ans;
};
int
main()
{
int
n, np, nc, _m, _u, _v, _c, ans;
while
(scanf(
"
%d%d%d%d
"
,
&
n,
&
np,
&
nc,
&
_m)
==
4
)
{
S
=
n, T
=
n
+
1
;
m
=
ans
=
0
;
memset(v,
-
1
,
sizeof
(v));
while
(_m
--
)
{
scanf(
"
%*[^(](%d,%d)%d
"
,
&
_u,
&
_v,
&
_c);
if
(_u
==
_v)
continue
;
insert(_u, _v, _c);
}
while
(np
--
)
{
scanf(
"
%*[^(](%d)%d
"
,
&
_v,
&
_c);
insert(S, _v, _c);
}
while
(nc
--
)
{
scanf(
"
%*[^(](%d)%d
"
,
&
_u,
&
_c);
insert(_u, T, _c);
}
printf(
"
%d\n
"
, dinic());
}
return
0
;
}