POJ 1984 Navigation Nightmare (并查集)
题意：给定一组线段和方向，然后查询点的距离。
思路：并查集的基本操作，在记录坐标偏移的时候注意一下，两个点之间和他们的根之间的坐标偏移关系可以用关系式表达出来，只要在纸上写一写就ok。查询输入的时候还是按ind排下序在处理线段。
/*
* 简单题 并查集的使用，就当是复习了。
*/
#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
=
40005
;
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))
typedef
struct
{
int
x,y;
}Node;
typedef
struct
{
int
a,b;
int
x,y;
}Road;
typedef
struct
{
int
a,b;
int
ind;
}Ask;
bool
cmp(
const
Ask
&
a,
const
Ask
&
b)
{
return
a.ind
<
b.ind;
}
Node node[MAXN];
Road road[MAXN
*
1000
];
Ask qus[MAXN
*
10
];
int
n,m,k;
int
pre[MAXN];
int
init()
{
for
(
int
i
=
0
; i
<
MAXN;
++
i)
node[i].x
=
node[i].y
=
0
;
CLR(pre,
-
1
);
return
0
;
}
int
find_set(
int
x)
{
int
root
=
x;
int
c_px,c_py,p_py,p_px;
int
tmp
=
0
;
p_px
=
node[x].x;
p_py
=
node[x].y;
while
(pre[root]
>=
0
)
{
root
=
pre[root];
node[x].x
+=
node[root].x;
node[x].y
+=
node[root].y;
}
while
( x
!=
root)
{
tmp
=
pre[x];
c_px
=
p_px;
c_py
=
p_py;
p_px
=
node[tmp].x;
p_py
=
node[tmp].y;
node[tmp].x
=
node[x].x
-
c_px;
node[tmp].y
=
node[x].y
-
c_py;
pre[x]
=
root;
x
=
tmp;
}
return
root;
}
void
union_set(
const
int
&
root1,
const
int
&
root2,
const
int
&
x,
const
int
&
y)
{
pre[root2]
+=
pre[root1];
pre[root1]
=
root2;
node[root1].x
=
x;
node[root1].y
=
y;
return
;
}
void
print()
{
for
(
int
i
=
1
; i
<=
n;
++
i)
printf(
"
(%d,%d)
"
,node[i].x,node[i].y);
printf(
"
%\n
"
);
}
int
input()
{
int
root1,root2,x,y,i,j,a,b,len;
char
c;
for
(i
=
0
; i
<
m;
++
i)
{
scanf(
"
%d %d %d %c
"
,
&
road[i].a,
&
road[i].b,
&
len,
&
c);
x
=
y
=
0
;
if
(c
==
'
N
'
)
y
=
len;
else
if
(c
==
'
S
'
)
y
=
-
len;
else
if
(c
==
'
W
'
)
x
=
-
len;
else
x
=
len;
road[i].x
=
x;
road[i].y
=
y;
}
IN(k);
for
(i
=
0
; i
<
k;
++
i)
scanf(
"
%d%d%d
"
,
&
qus[i].a,
&
qus[i].b,
&
qus[i].ind);
return
0
;
}
int
work()
{
int
i,j,tmp,root1,root2,ans,a,b,t,x,y;
int
pre
=
0
;
sort(qus,qus
+
k,cmp);
for
(i
=
0
; i
<
k;
++
i)
{
a
=
qus[i].a;
b
=
qus[i].b;
t
=
qus[i].ind;
for
(; pre
<
t;
++
pre)
{
root1
=
find_set(road[pre].a);
root2
=
find_set(road[pre].b);
if
(root1
==
root2)
continue
;
x
=
road[pre].x
-
node[road[pre].a].x
+
node[road[pre].b].x;
y
=
road[pre].y
-
node[road[pre].a].y
+
node[road[pre].b].y;
union_set(root1,root2,x,y);
}
root1
=
find_set(a);
root2
=
find_set(b);
ans
=
-
1
;
if
(root1
!=
root2)
OUT(ans);
else
{
root1
=
ABS(node[a].x
-
node[b].x);
root2
=
ABS(node[a].y
-
node[b].y);
ans
=
root1
+
root2;
OUT(ans);
}
}
return
0
;
}
int
main()
{
while
(scanf(
"
%d%d
"
,
&
n,
&
m)
!=
EOF)
{
init();
input();
work();
}
return
0
;
}