二分图最大匹配
HDU 1150:
http://acm.hdu.edu.cn/showproblem.php?pid=1150
最小覆盖点 == 最大匹配数(选取最少的点数，使这些点和所有的边都有关联——把所有的边的覆盖)
两台机器，有n和m个工作模式，起始工作模式都为0，现在有k件工作，第i件工作可分别在两个机器上用各自的模式工作，但换模式要重启，问重启的最小次数。
写的时候因为是找二分最大匹配的题目时找到写的，想到了二分上去，也知道是求最小覆盖点 == 最大匹配数，但不是很能理解，先把代码写了再说。
写的时候注意起始模式是0，所以换模式时把0的排除再外。（因为这个原因错了很多次）
一：邻接阵做法
代码
#include
<
iostream
>
using
namespace
std;
int
n,m,k;
int
map[
105
][
105
];
//
记录X,Y对应点可否连接
int
vis[
105
];
//
每次找增广路时对Y中点是否访问
int
dir[
105
];
//
Y中点匹配的X中点的位置
int
find(
int
a)
{
int
i;
for
(i
=
0
;i
<
m;i
++
)
{
if
(map[a][i]
==
1
&&
vis[i]
==
0
)
{
vis[i]
=
1
;
if
(dir[i]
==
-
1
||
find(dir[i]))
{
dir[i]
=
a;
return
1
;
}
}
}
return
0
;
}
int
main()
{
int
i,x,y;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF
&&
n
!=
0
)
{
memset(map,
0
,
sizeof
(map));
memset(dir,
-
1
,
sizeof
(dir));
scanf(
"
%d%d
"
,
&
m,
&
k);
while
(k
--
)
{
scanf(
"
%d%d%d
"
,
&
i,
&
x,
&
y);
if
(x
&&
y)
map[x][y]
=
1
;
//
应该不能再用map[y][x] = 1
}
int
sum
=
0
;
for
(i
=
0
;i
<
n;i
++
)
{
memset(vis,
0
,
sizeof
(vis));
if
(find(i))
sum
++
;
}
printf(
"
%d\n
"
,sum);
}
return
0
;
}
二：动态邻接表做法.
代码
#include
<
iostream
>
using
namespace
std;
int
n,m,k;
int
vis[
105
];
//
每次找增广路时对Y中点是否访问
int
dir[
105
];
//
Y中点匹配的X中点的位置
struct
edge
{
int
from;
int
to;
edge
*
next;
edge()
{
from
=
to
=
0
;
next
=
NULL;
}
};
edge
*
List[
105
];
void
add_edge(
int
f,
int
t)
{
edge
*
node
=
new
edge();
node
->
from
=
f;
node
->
to
=
t;
node
->
next
=
List[f];
List[f]
=
node;
}
int
find(edge
*
node)
{
while
(
1
)
{
if
(node
==
NULL)
{
break
;
}
int
i
=
node
->
to;
if
(vis[i]
==
0
)
{
vis[i]
=
1
;
if
(dir[i]
==
-
1
||
find(List[dir[i]]))
{
dir[i]
=
node
->
from;
return
1
;
}
}
node
=
node
->
next;
}
return
0
;
}
int
main()
{
int
i,x,y;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF
&&
n
!=
0
)
{
for
(i
=
0
;i
<=
n;i
++
)
//
链表清空，一开始没清空，错了很多次
{
List[i]
=
NULL;
}
memset(dir,
-
1
,
sizeof
(dir));
scanf(
"
%d%d
"
,
&
m,
&
k);
while
(k
--
)
{
scanf(
"
%d%d%d
"
,
&
i,
&
x,
&
y);
if
(x
&&
y)
add_edge(x,y);
}
int
sum
=
0
;
for
(i
=
1
;i
<=
n;i
++
)
{
memset(vis,
0
,
sizeof
(vis));
if
(find(List[i]))
sum
++
;
}
printf(
"
%d\n
"
,sum);
}
return
0
;
}
HDU 1151:
http://acm.hdu.edu.cn/showproblem.php?pid=1151
最小路径覆盖（选取最少的边覆盖所有的点） == 节点数 - 最大匹配数
代码
#include
<
iostream
>
using
namespace
std;
#define
MAXN 125
int
vis[MAXN];
int
link[MAXN];
int
n,m;
struct
edge
{
int
from;
int
to;
edge
*
next;
edge()
{
from
=
to
=
0
;
next
=
NULL;
}
};
edge
*
List[MAXN];
void
add_edge(
int
f,
int
t)
{
edge
*
node
=
new
edge();
node
->
from
=
f;
node
->
to
=
t;
node
->
next
=
List[f];
List[f]
=
node;
}
bool
find(edge
*
node)
{
while
(
1
)
{
if
(node
==
NULL)
break
;
int
i
=
node
->
to;
if
(vis[i]
==
0
)
{
vis[i]
=
1
;
if
(link[i]
==
0
||
find(List[link[i]]))
{
link[i]
=
node
->
from;
return
1
;
}
}
node
=
node
->
next;
}
return
0
;
}
int
main()
{
int
i;
int
t,x,y;
scanf(
"
%d
"
,
&
t);
while
(t
--
)
{
memset(link,
0
,
sizeof
(link));
scanf(
"
%d%d
"
,
&
n,
&
m);
for
(i
=
0
;i
<=
n;i
++
)
{
List[i]
=
NULL;
}
for
(i
=
0
;i
<
m;i
++
)
{
scanf(
"
%d%d
"
,
&
x,
&
y);
add_edge(x,y);
}
int
sum
=
0
;
for
(i
=
1
;i
<=
n;i
++
)
{
memset(vis,
0
,
sizeof
(vis));
if
(find(List[i]))
sum
++
;
}
printf(
"
%d\n
"
,n
-
sum);
}
return
0
;
}
HDU 1068:
http://acm.hdu.edu.cn/showproblem.php?pid=1068
二分图最大独立集=顶点数-二分图最大匹配
代码
#include
<
iostream
>
using
namespace
std;
const
long
max_edge
=
100005
;
const
long
max_point
=
1005
;
int
n;
int
Index;
struct
node
{
int
y;
int
next;
}stu[max_edge];
int
pre[max_point];
//
以该点为起点的第一条在stu中的存储位置
int
vis[max_point];
//
Y中结点的访问情况
int
link[max_point];
//
Y中与X中配对的情况
void
add_edge(
int
a,
int
b)
{
stu[Index].y
=
b;
stu[Index].next
=
pre[a];
pre[a]
=
Index
++
;
}
void
Init()
{
int
i,j,s,t,m;
char
c;
Index
=
1
;
memset(pre,
-
1
,
sizeof
(pre));
memset(link,
-
1
,
sizeof
(link));
for
(i
=
0
;i
<
n;i
++
)
{
cin
>>
s
>>
c
>>
c
>>
m
>>
c;
for
(j
=
0
;j
<
m;j
++
)
{
scanf(
"
%d
"
,
&
t);
add_edge(s,t);
}
}
}
bool
find(
int
dir)
{
int
i;
for
(i
=
pre[dir]; i
!=-
1
; i
=
stu[i].next)
{
int
ii
=
stu[i].y;
if
(vis[ii]
==
0
)
{
vis[ii]
=
1
;
if
(link[ii]
==
-
1
||
find(link[ii]))
{
link[ii]
=
dir;
return
true
;
}
}
}
return
false
;
}
void
Play()
{
int
i;
int
ans
=
0
;
for
(i
=
0
;i
<
n;i
++
)
{
memset(vis,
0
,
sizeof
(vis));
if
(find(i))
ans
++
;
}
printf(
"
%d\n
"
,n
-
ans
/
2
);
}
int
main()
{
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
Init();
Play();
}
return
0
;
}
HDU 1281:
http://acm.hdu.edu.cn/showproblem.php?pid=1281
写的时候写搓了，一个小错误，find1递归的时候用了find函数，很是郁闷!
当数据量小的时候可以不用静态连接表存储，毕竟连接阵用起来比连接表方便!
代码
#include
<
iostream
>
using
namespace
std;
const
long
max_point
=
1005
;
const
long
max_edge
=
100005
;
int
n,m,k;
int
cas
=
0
;
int
mat[max_point][max_point];
int
vis[max_point];
int
linky[max_point];
int
linkx[max_point];
void
Init()
{
int
i;
int
a,b;
memset(mat,
0
,
sizeof
(mat));
for
(i
=
0
;i
<
k;i
++
)
{
scanf(
"
%d%d
"
,
&
a,
&
b);
mat[a][b]
=
1
;
}
}
bool
find(
int
dir,
bool
flag)
{
int
i;
for
(i
=
1
;i
<=
m;i
++
)
{
if
(mat[dir][i]
&&
!
vis[i])
{
vis[i]
=
1
;
if
(linky[i]
==-
1
||
find(linky[i],flag))
{
if
(flag)
{
linkx[dir]
=
i;
linky[i]
=
dir;
}
return
true
;
}
}
}
return
false
;
}
bool
can()
{
int
i;
for
(i
=
1
;i
<=
n;i
++
)
{
if
(linkx[i]
==
-
1
)
{
memset(vis,
0
,
sizeof
(vis));
if
(find(i,
0
))
{
return
true
;
}
}
}
return
false
;
}
void
Play()
{
int
i,j;
int
ans
=
0
;
memset(linkx,
-
1
,
sizeof
(linkx));
memset(linky,
-
1
,
sizeof
(linky));
for
(i
=
1
;i
<=
n;i
++
)
{
memset(vis,
0
,
sizeof
(vis));
if
(find(i,
1
))
ans
++
;
}
int
ans1
=
0
;
for
(i
=
1
;i
<=
n;i
++
)
{
if
(linkx[i]
!=
-
1
)
{
int
tmp
=
linkx[i];
linkx[i]
=
-
1
;
linky[tmp]
=
-
1
;
mat[i][tmp]
=
0
;
if
(
!
can())
ans1
++
;
mat[i][tmp]
=
1
;
linky[tmp]
=
i;
linkx[i]
=
tmp;
}
}
printf(
"
Board %d have %d important blanks for %d chessmen.\n
"
,
++
cas,ans1,ans);
}
int
main()
{
while
(scanf(
"
%d%d%d
"
,
&
n,
&
m,
&
k)
!=
EOF)
{
Init();
Play();
}
return
0
;
}
HDU 1498:
http://acm.hdu.edu.cn/showproblem.php?pid=1498
感觉来了，这题没花什么精力，一杯咖啡的时候想了出来。对每个颜色，求最小点覆盖，如果大于k，则不能在k时间内完成，输出；否则可以完成。
代码
#include
<
iostream
>
using
namespace
std;
int
n,k;
const
long
max_color
=
55
;
const
long
maxn
=
105
;
int
Hash[max_color];
int
map[max_color][maxn][maxn];
int
vis[maxn];
int
linkx[maxn],linky[maxn];
void
Init()
{
int
i,j;
memset(map,
0
,
sizeof
(map));
memset(Hash,
0
,
sizeof
(Hash));
for
(i
=
0
;i
<
n;i
++
)
{
for
(j
=
0
;j
<
n;j
++
)
{
int
color;
scanf(
"
%d
"
,
&
color);
map[color][i][j]
=
1
;
Hash[color]
=
1
;
}
}
}
bool
find(
int
dir,
int
u)
{
int
i,j;
for
(i
=
0
;i
<
n;i
++
)
{
if
(
1
==
map[dir][u][i])
{
if
(
0
==
vis[i])
{
vis[i]
=
1
;
if
(linky[i]
==
-
1
||
find(dir,linky[i]))
{
linky[i]
=
u;
linkx[u]
=
i;
return
true
;
}
}
}
}
return
false
;
}
int
MaxMatch(
int
dir)
{
int
i;
int
sum
=
0
;
memset(linkx,
-
1
,
sizeof
(linkx));
memset(linky,
-
1
,
sizeof
(linky));
for
(i
=
0
;i
<
n;i
++
)
{
if
(
-
1
==
linkx[i])
{
memset(vis,
0
,
sizeof
(vis));
if
(find(dir,i))
{
sum
++
;
}
}
}
return
sum;
}
void
Play()
{
int
i;
int
flag
=
0
;
for
(i
=
1
;i
<=
50
;i
++
)
{
//
对应每种颜色
if
(
1
==
Hash[i])
{
int
num
=
MaxMatch(i);
/*
if(flag != 0)
{//这个放在外面，PE了一次
printf(" ");
}
*/
if
(num
>
k)
{
if
(flag
!=
0
)
{
printf(
"
"
);
}
printf(
"
%d
"
,i);
flag
=
1
;
}
}
}
if
(
0
==
flag)
{
printf(
"
-1
"
);
}
printf(
"
\n
"
);
}
int
main()
{
while
(scanf(
"
%d%d
"
,
&
n,
&
k)
!=
EOF
&&
(n
!=
0
||
k
!=
0
))
{
Init();
Play();
}
return
0
;
}
HDU 1528:
http://acm.hdu.edu.cn/showproblem.php?pid=1528
简单题，卡片为点，大小关系为边，求最大匹配。
代码
#include
<
iostream
>
#include
<
map
>
using
namespace
std;
const
long
maxn
=
30
;
int
k;
map
<
char
,
int
>
M;
int
num_Adam[maxn],num_Eve[maxn];
int
linkx[maxn],linky[maxn];
int
mat[maxn][maxn],vis[maxn],pre[maxn];
void
Init()
{
int
i,j;
char
c1,c2;
memset(mat,
0
,
sizeof
(mat));
M[
'
2
'
]
=
0
;M[
'
3
'
]
=
1
;M[
'
4
'
]
=
2
;M[
'
5
'
]
=
3
;M[
'
6
'
]
=
4
;
M[
'
7
'
]
=
5
;M[
'
8
'
]
=
6
;M[
'
9
'
]
=
7
;M[
'
T
'
]
=
8
;M[
'
J
'
]
=
9
;
M[
'
Q
'
]
=
10
;M[
'
K
'
]
=
11
;M[
'
A
'
]
=
12
;
M[
'
H
'
]
=
3
;M[
'
S
'
]
=
2
;M[
'
D
'
]
=
1
;M[
'
C
'
]
=
0
;
scanf(
"
%d
"
,
&
k);
for
(i
=
0
;i
<
k;i
++
)
{
cin
>>
c1
>>
c2;
num_Adam[i]
=
M[c1]
*
4
+
M[c2];
}
for
(i
=
0
;i
<
k;i
++
)
{
cin
>>
c1
>>
c2;
num_Eve[i]
=
M[c1]
*
4
+
M[c2];
}
for
(i
=
0
;i
<
k;i
++
)
{
for
(j
=
0
;j
<
k;j
++
)
{
if
(num_Adam[i]
<
num_Eve[j])
{
mat[i][j]
=
1
;
}
}
}
}
bool
find(
int
dir)
{
int
i;
for
(i
=
0
;i
<
k;i
++
)
{
if
(
1
==
mat[dir][i]
&&
0
==
vis[i])
{
vis[i]
=
1
;
if
(linky[i]
==
-
1
||
find(linky[i]))
{
linky[i]
=
dir;
linkx[dir]
=
i;
return
true
;
}
}
}
return
false
;
}
void
Play()
{
int
i;
int
ans
=
0
;
memset(linkx,
-
1
,
sizeof
(linkx));
memset(linky,
-
1
,
sizeof
(linky));
for
(i
=
0
;i
<
k;i
++
)
{
if
(
-
1
==
linkx[i])
{
memset(vis,
0
,
sizeof
(vis));
if
(find(i))
ans
++
;
}
}
printf(
"
%d\n
"
,ans);
}
int
main()
{
int
t;
scanf(
"
%d
"
,
&
t);
while
(t
--
)
{
Init();
Play();
}
return
0
;
}