POJ 2584 T-Shirt Gumbo(最大流)
题意：有n个同学要求订购衣服，每个同学可以接收的衣服型号已知（一个人一件衣服），现在每个型号的衣服数量已知，问能不能满足同学的要求？
思路：可行性问题，此题方法很多，多重二分匹配、BFS等，我偷懒用了最大流，BS我吧.......还是说下建图过程吧，增加源汇点s,t，每个同学与其所接受的型号连一条1的边，s与同学连INF边，型号和t连val (val 是此型号的数量)，然后求最大流......
#include
<
iostream
>
#include
<
cstdio
>
#include
<
algorithm
>
#include
<
memory.h
>
#include
<
cmath
>
#include
<
bitset
>
#include
<
queue
>
#include
<
vector
>
using
namespace
std;
const
int
BORDER
=
(
1
<<
20
)
-
1
;
const
int
MAXSIZE
=
37
;
const
int
MAXN
=
200
;
const
int
INF
=
INT_MAX;
#define
CLR(x,y) memset(x,y,sizeof(x))
#define
ADD(x) x=((x+1)&BORDER)
#define
IN(x) scanf("%d",&x)
#define
OUT(x) printf("%d\n",x)
#define
MIN(m,v) (m)<(v)?(m):(v)
#define
MAX(m,v) (m)>(v)?(m):(v)
#define
ABS(x) ((x)>0?(x):-(x))
int
index,nv,n,m,k,s,t;
int
ans;
int
net[MAXN];
int
need[
30
];
int
have[
10
];
struct
Edge{
int
next,pair;
int
v,cap,flow;
}edge[MAXN
*
MAXN];
int
init()
{
CLR(net,
-
1
);
CLR(have,
0
);
CLR(need,
0
);
index
=
0
;
need[
'
S
'
]
=
1
;
need[
'
M
'
]
=
2
;
need[
'
L
'
]
=
3
;
need[
'
X
'
]
=
4
;
need[
'
T
'
]
=
5
;
return
0
;
}
void
add_edge(
const
int
&
u,
const
int
&
v,
const
int
&
val)
{
edge[index].next
=
net[u];
net[u]
=
index;
edge[index].v
=
v;
edge[index].cap
=
val;
edge[index].flow
=
0
;
edge[index].pair
=
index
+
1
;
++
index;
edge[index].next
=
net[v];
net[v]
=
index;
edge[index].v
=
u;
edge[index].cap
=
0
;
edge[index].flow
=
0
;
edge[index].pair
=
index
-
1
;
++
index;
}
int
input()
{
string
op;
int
i,j,start,end;
cin
>>
op;
if
(op.length()
>
6
)
return
1
;
cin
>>
n;
s
=
n
+
5
+
1
;
t
=
n
+
5
+
2
;
nv
=
t;
for
(i
=
1
; i
<=
n;
++
i)
{
cin
>>
op;
for
(j
=
need[op[
0
]]; j
<=
need[op[
1
]];
++
j)
add_edge(i,n
+
j,
1
);
add_edge(s,i,
1
);
}
for
(i
=
1
; i
<=
5
;
++
i)
{
cin
>>
j;
add_edge(i
+
n,t,j);
}
cin
>>
op;
return
0
;
}
int
ISAP()
{
long
numb[MAXN],dist[MAXN],curedge[MAXN],pre[MAXN];
long
cur_flow,max_flow,u,tmp,neck,i;
memset(dist,
0
,
sizeof
(dist));
memset(numb,
0
,
sizeof
(numb));
for
(i
=
1
; i
<=
nv ;
++
i)
curedge[i]
=
net[i];
numb[nv]
=
nv;
max_flow
=
0
;
u
=
s;
while
(dist[s]
<
nv)
{
/*
first , check if has augmemt flow
*/
if
(u
==
t)
{
cur_flow
=
INF;
for
(i
=
s; i
!=
t;i
=
edge[curedge[i]].v)
{
if
(cur_flow
>
edge[curedge[i]].cap)
{
neck
=
i;
cur_flow
=
edge[curedge[i]].cap;
}
}
for
(i
=
s; i
!=
t; i
=
edge[curedge[i]].v)
{
tmp
=
curedge[i];
edge[tmp].cap
-=
cur_flow;
edge[tmp].flow
+=
cur_flow;
tmp
=
edge[tmp].pair;
edge[tmp].cap
+=
cur_flow;
edge[tmp].flow
-=
cur_flow;
}
max_flow
+=
cur_flow;
u
=
s;
}
/*
if .... else ...
*/
for
(i
=
curedge[u]; i
!=
-
1
; i
=
edge[i].next)
if
(edge[i].cap
>
0
&&
dist[u]
==
dist[edge[i].v]
+
1
)
break
;
if
(i
!=
-
1
)
{
curedge[u]
=
i;
pre[edge[i].v]
=
u;
u
=
edge[i].v;
}
else
{
if
(
0
==
--
numb[dist[u]])
break
;
curedge[u]
=
net[u];
for
(tmp
=
nv,i
=
net[u]; i
!=
-
1
; i
=
edge[i].next)
if
(edge[i].cap
>
0
)
tmp
=
tmp
<
dist[edge[i].v]
?
tmp:dist[edge[i].v];
dist[u]
=
tmp
+
1
;
++
numb[dist[u]];
if
(u
!=
s) u
=
pre[u];
}
}
return
max_flow;
}
int
work()
{
int
i,j,tmp;
ans
=
ISAP();
if
(ans
==
n)
printf(
"
T-shirts rock!\n
"
);
else
printf(
"
I'd rather not wear a shirt anyway...\n
"
);
return
0
;
}
int
main()
{
int
res;
while
(
true
)
{
init();
res
=
input();
if
(res)
break
;
work();
}
return
0
;
}