POJ 2289 Jamie's Contact Groups (二分答案+匹配流)
题意：Jamie有很多联系人，但是很不方便管理，他想把这些联系人分成组，已知这些联系人可以被分到哪个组中去，而且要求每个组的联系人上限最小，即有一整数k，使每个组的联系人数都不大于k，问这个k最小是多少？
思路：这个题很像以前的那个星际战舰的那个题，一般就是二分答案，对每一个值进行建图，求最大流，查看这个最大流是否等于总人数，虽然比较繁琐，但是思路还是很清晰的，之后就是SAP模板了......(每次建图的时候一定要对信息进行初始化啊！！！WA了N次！)
#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
=
1508
;
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,n,m,ans;
int
adj[MAXN][MAXN],
in
[MAXN];
int
nv,s,t;
int
net[MAXN];
char
str[
20
];
struct
Edge{
int
next,pair;
int
v,cap,flow;
}edge[
1500000
];
int
init()
{
index
=
0
;
CLR(net,
-
1
);
CLR(adj,
0
);
CLR(
in
,
0
);
ans
=
0
;
return
0
;
}
int
input()
{
int
i,j,tmp;
char
ch;
getchar();
for
(i
=
1
;i
<=
n; i
++
)
{
ch
=
getchar();
while
((ch
>=
'
a
'
&&
ch
<=
'
z
'
)
||
(ch
>=
'
A
'
&&
ch
<=
'
Z
'
))
ch
=
getchar();
while
(ch
==
'
'
)
ch
=
getchar();
while
(
true
)
{
j
=
0
;
while
(ch
>=
'
0
'
&&
ch
<=
'
9
'
)
{
j
=
j
*
10
+
ch
-
'
0
'
;
ch
=
getchar();
}
adj[i][j
+
1
]
=
1
;
in
[j
+
1
]
++
;
if
(ch
==
'
\n
'
)
break
;
while
(ch
==
'
'
) ch
=
getchar();
}
}
s
=
0
;
nv
=
n
+
m
+
2
;
t
=
nv
-
1
;
return
1
;
}
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
make_graph(
const
int
&
limit)
{
int
i,j,tmp,k,u,v,val;
CLR(net,
-
1
);
index
=
0
;
for
(i
=
1
; i
<=
n;
++
i)
add_edge(s,i,
1
);
for
(i
=
1
; i
<=
n;
++
i)
{
for
(j
=
1
; j
<=
m;
++
j)
if
(adj[i][j])
add_edge(i,j
+
n,
1
);
}
for
(i
=
1
; i
<=
m;
++
i)
if
(
in
[i]
>=
limit)
add_edge(i
+
n,t,limit);
else
add_edge(i
+
n,t,
in
[i]);
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
low,high,mid,tmp;
int
pre;
low
=
0
;
high
=
-
1
;
for
(
int
i
=
1
; i
<=
m;
++
i)
high
=
MAX(high,
in
[i]);
while
(low
<=
high)
{
mid
=
(low
+
high)
>>
1
;
make_graph(mid);
tmp
=
ISAP();
if
(tmp
==
n)
{
pre
=
mid;
high
=
mid
-
1
;
}
else
low
=
mid
+
1
;
}
printf(
"
%d\n
"
,pre);
return
0
;
}
int
main()
{
while
(scanf(
"
%d%d
"
,
&
n,
&
m)
!=
EOF)
{
if
(
!
n
&&
!
m)
break
;
init();
input();
work();
}
return
0
;
}