PKU 2446
同3020，关键是建图，对每个点的四个相邻点，根据是否有hole，建立边的关系。
代码
//
Rank:40 0ms
#include
<
stdio.h
>
#include
<
string
.h
>
#define
NL 35
#define
SL NL*NL
int
adj[SL][
5
];
int
mth[SL];
bool
used[SL];
bool
mx[NL][NL];
int
flg[NL][NL];
int
dir[
4
][
2
]
=
{{
0
,
1
}, {
0
,
-
1
}, {
1
,
0
}, {
-
1
,
0
}};
bool
path(
int
u)
{
for
(
int
v
=
1
; v
<=
adj[u][
0
]; v
++
) {
int
t
=
adj[u][v];
if
(
!
used[t]) {
used[t]
=
1
;
if
(
!
mth[t]
||
path(mth[t])) {
mth[t]
=
u;
return
1
;
}
}
}
return
0
;
}
int
main()
{
int
m, n, k;
int
i, j, r, c, nt;
while
(scanf(
"
%d%d%d
"
,
&
m,
&
n,
&
k)
!=
EOF) {
memset(mx,
0
,
sizeof
(mx));
for
(i
=
0
; i
<
k; i
++
) {
scanf(
"
%d%d
"
,
&
c,
&
r);
mx[r][c]
=
1
;
}
if
((m
*
n
-
k)
&
1
) {
printf(
"
NO\n
"
);
continue
;
}
memset(adj,
0
,
sizeof
(adj));
memset(flg,
0
,
sizeof
(flg));
nt
=
0
;
for
(i
=
1
; i
<=
m; i
++
) {
for
(j
=
1
; j
<=
n; j
++
) {
if
(
!
mx[i][j]) {
flg[i][j]
=
++
nt;
}
}
}
for
(i
=
1
; i
<=
m; i
++
) {
for
(j
=
1
; j
<=
n; j
++
) {
if
(
!
mx[i][j]) {
for
(
int
t
=
0
; t
<
4
; t
++
) {
int
x
=
i
+
dir[t][
0
];
int
y
=
j
+
dir[t][
1
];
if
(x
>=
1
&&
x
<=
m
&&
y
>=
1
&&
y
<=
n
&&
!
mx[x][y]) {
int
f1
=
flg[i][j];
int
f2
=
flg[x][y];
adj[f1][
++
adj[f1][
0
]]
=
f2;
}
}
}
}
}
int
ans
=
0
;
memset(mth,
0
,
sizeof
(mth));
for
(i
=
1
; i
<=
nt; i
++
) {
memset(used,
0
,
sizeof
(used));
if
(path(i)) ans
++
;
}
if
(ans
==
m
*
n
-
k) {
printf(
"
YES\n
"
);
}
else
{
printf(
"
NO\n
"
);
}
}
return
0
;
}