POJ 2922 Honeymoon Hike (dfs+二分答案+区间枚举)
题意：给定一个n*n地图和高度，一个人从左上角地图开始，需要走到右下角的地图，问他走的路程中所到达的最大高度和最小高度的差最小是多少。
思路：二分答案，对于每一个高度差，可以用dfs或bfs检验是否能够到达目的地，如果对每个高度差，用dfs，每次传以前的最大和最小高度，然后在更新，这样检验会超时，于是可以改用枚举高度区间的办法来缩减时间(想了半天没想出来....)。总的来所，一个不错的搜索题目。
#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
=
105
;
const
int
INF
=
1000000000
;
#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))
#define
SET_NODE(no,a,b) {no.x=a;no.y=b;}
typedef
struct
{
int
x,y;
}Node;
Node t_node;
int
dir[
4
][
2
]
=
{{
1
,
0
},{
0
,
1
},{
-
1
,
0
},{
0
,
-
1
}};
int
n,tt,ans,limit,low,high;
int
map[MAXN][MAXN];
bool
visit[MAXN][MAXN];
bool
_is(
const
int
&
x,
const
int
&
y)
{
if
(x
<
0
||
x
>=
n)
return
false
;
if
(y
<
0
||
y
>=
n)
return
false
;
return
true
;
}
int
init()
{
CLR(visit,
0
);
return
0
;
}
int
input()
{
IN(n);
for
(
int
i
=
0
; i
<
n;
++
i)
for
(
int
j
=
0
; j
<
n;
++
j)
IN(map[i][j]);
return
0
;
}
/*
bool bfs(const int& limit)
{
int x,y,i,j,tmp,high,low;
queue<Node> que;
Node node;
CLR(visit,0);
// initialize the queue
while(!que.empty())
que.pop();
visit[0][0] = true;
SET_NODE(t_node,0,0);
que.push(t_node);
low = high = map[0][0];
// while loop
cout<<limit<<endl;
while(!que.empty())
{
node = que.front();
que.pop();
if(node.x == n-1 && node.y == n-1)
return true;
for(i = 0; i < 4; ++i)
{
x = node.x + dir[i][0];
y = node.y + dir[i][1];
if(_is(x,y) && !visit[x][y])
{
t_node.low = MIN(low,map[x][y]);
t_node.high = MAX(high,map[x][y]);
if(high - low > limit)
continue;
SET_NODE(t_node,x,y);
visit[x][y] = true;
que.push(t_node);
}
}
}
return false;
}
*/
bool
dfs(
const
int
&
x,
const
int
&
y)
{
int
a,b,h,l;
if
(map[x][y]
>
high
||
map[x][y]
<
low)
return
false
;
if
(x
==
n
-
1
&&
y
==
n
-
1
)
return
true
;
visit[x][y]
=
true
;
for
(
int
i
=
0
; i
<
4
;
++
i)
{
a
=
x
+
dir[i][
0
];
b
=
y
+
dir[i][
1
];
if
(
!
visit[a][b]
&&
_is(a,b))
if
(dfs(a,b))
return
true
;
}
return
false
;
}
int
work()
{
int
b_high,b_low,b_mid;
int
i,j,tmp;
b_high
=
-
1
;
b_low
=
INF;
for
(i
=
0
; i
<
n;
++
i)
for
(j
=
0
; j
<
n;
++
j)
{
b_high
=
MAX(b_high,map[i][j]);
b_low
=
MIN(b_low,map[i][j]);
}
b_high
-=
b_low;
b_low
=
0
;
while
(b_low
<=
b_high)
{
b_mid
=
(b_low
+
b_high)
>>
1
;
for
(low
=
(map[
0
][
0
]
-
b_mid)
>
0
?
(map[
0
][
0
]
-
b_mid):
0
;
low
<=
map[
0
][
0
];
++
low)
{
CLR(visit,
0
);
high
=
low
+
b_mid;
if
(dfs(
0
,
0
))
{
ans
=
b_mid;
break
;
}
}
if
(low
<=
map[
0
][
0
])
b_high
=
b_mid
-
1
;
else
b_low
=
b_mid
+
1
;
}
printf(
"
Scenario #%d:\n
"
,tt);
OUT(ans);
printf(
"
\n
"
);
tt
++
;
return
0
;
}
int
main()
{
int
t;
tt
=
1
;
IN(t);
while
(t
--
)
{
init();
input();
work();
}
return
0
;
}