ZOJ Problem Set - 2850
Beautiful Meadow
Time Limit: 1 Second
Memory Limit: 32768 KB
Tom's Meadow
Tom has a meadow in his garden. He divides it into N * M squares. Initially all the squares were covered with grass. He mowed down the grass on some of the squares and thinks the meadow is beautiful if and only if
Not all squares are covered with grass.No two mowed squares are adjacent.
Two squares are adjacent if they share an edge. Here comes the problem: Is Tom's meadow beautiful now?
Input
The input contains multiple test cases!
Each test case starts with a line containing two integers N, M (1 <= N, M <= 10) separated by a space. There follows the description of Tom's Meadow. There're N lines each consisting of M integers separated by a space. 0(zero) means the corresponding position of the meadow is mowed and 1(one) means the square is covered by grass.
A line with N = 0 and M = 0 signals the end of the input, which should not be processed
Output
One line for each test case.
Output "Yes" (without quotations) if the meadow is beautiful, otherwise "No"(without quotations).
Sample Input
2 2 1 0 0 1 2 2 1 1 0 0 2 3 1 1 1 1 1 1 0 0
Sample Output
Yes No No
Author: CAO, Peng
Source:
Zhejiang Provincial Programming Contest 2007
Submit
Status
//
1859941 2009-05-08 14:40:12 Accepted 2850 C++ 0 184 Wpl
//
1859935 2009-05-08 14:29:21 Wrong Answer 2850 C++ 0 184 Wpl
#include
<
iostream
>
#define
MAX 12
using
namespace
std;
int
map[MAX][MAX];
int
a[
4
][
2
]
=
{{
0
,
1
},{
0
,
-
1
},{
1
,
0
},{
-
1
,
0
}};
int
n,m,sum;
bool
mark;
void
Init() {
int
i,j;
for
(i
=
0
;i
<
n;i
++
)
for
(j
=
0
;j
<
m;j
++
) scanf(
"
%d
"
,
&
map[i][j]); }
bool
Bound(
int
x,
int
y) {
if
(x
>=
0
&&
y
>=
0
&&
x
<
n
&&
y
<
m)
return
true
;
else
return
false
; }
void
Decide() {
int
i,j,ti,tj,k;
for
(i
=
0
;i
<
n;i
++
)
for
(j
=
0
;j
<
m;j
++
) {
if
(map[i][j]
==
1
) { sum
++
;
continue
; }
for
(k
=
0
;k
<
4
;k
++
) { ti
=
i
+
a[k][
0
];
//
此处小心,绝不能写成i或者k，这里错过两次了
tj
=
j
+
a[k][
1
];
if
(Bound(ti,tj)) {
if
(map[ti][tj]
==
0
) { mark
=
true
;
return
; } } } } }
int
main() {
while
(scanf(
"
%d%d
"
,
&
n,
&
m)
!=
EOF) {
if
(n
==
0
&&
m
==
0
)
break
; Init(); mark
=
false
; sum
=
0
; Decide();
if
(mark
||
sum
==
m
*
n) printf(
"
No\n
"
);
else
printf(
"
Yes\n
"
); }
return
0
; }
转载于:https://www.cnblogs.com/forever4444/archive/2009/05/08/1452640.html
