题目大意：自行脑补。
思路：维护两个维度上的带权并查集即可。
注意对于题目给出的一堆关系，我们应该添加两对关系。
Code:
#include
<cstdio>#include
<cstring>#include
<cctype>#include
<iostream>#include
<algorithm>using
namespace
std;#define
N 40010int
n, m;struct
UnionSet { int
root[N], dis[N]; void
reset() { int
i; for(i
= 1; i <= n; ++i) root[i]
= i, dis[i] = 0; } int
find(int
x) { static
int
stack[N]; int
top = 0; for(;
x != root[x]; x = root[x]) stack[++top]
= x; for(int
i = top - 1; i >= 1; --i) dis[stack[i]]
+= dis[stack[i + 1]], root[stack[i]] = x; return
x; }}Set[2];#define
K 10010struct
Ask { int
u, v, tclock, lab; void
read(int
_) { lab
= _; scanf("%d%d%d",
&u, &v, &tclock); } bool
operator < (const
Ask &B) const
{ return
tclock < B.tclock; }}S[K];int
ans[K];#define
M 40010int
u[M], v[M], d[M];bool
vec[M];#define
_abs(x) ((x)>0?(x):-(x))int
main() { #ifndef
ONLINE_JUDGE freopen("tt.in",
"r",
stdin); #endif
scanf("%d%d",
&n, &m); register
int
i, j; int
a, b, x; char
s[10]; for(i
= 1; i <= m; ++i) { scanf("%d%d%d%s",
&a, &b, &x, s); d[i]
= x; if
(s[0] == 'E') u[i]
= a, v[i] = b, vec[i] = 0; else
if
(s[0] == 'W') u[i]
= b, v[i] = a, vec[i] = 0; else
if
(s[0] == 'N') u[i]
= a, v[i] = b, vec[i] = 1; else u[i]
= b, v[i] = a, vec[i] = 1; } int
Q; scanf("%d",
&Q); for(i
= 1; i <= Q; ++i) S[i].read(i); sort(S
+ 1, S + Q + 1); Set[0].reset(),
Set[1].reset(); int
ra, rb; j
= 1; for(i
= 1; i <= m; ++i) { ra
= Set[vec[i]].find(u[i]); rb
= Set[vec[i]].find(v[i]); Set[vec[i]].root[ra]
= rb; Set[vec[i]].dis[ra]
= d[i] + Set[vec[i]].dis[v[i]] - Set[vec[i]].dis[u[i]]; ra
= Set[1 - vec[i]].find(u[i]); rb
= Set[1 - vec[i]].find(v[i]); Set[1
- vec[i]].root[ra] = rb; Set[1
- vec[i]].dis[ra] = Set[1 - vec[i]].dis[v[i]] - Set[1 - vec[i]].dis[u[i]]; for(;
S[j].tclock == i; ++j) { ra
= Set[0].find(S[j].u), rb = Set[0].find(S[j].v); if
(ra != rb) { ans[S[j].lab]
= -1; continue; } ans[S[j].lab]
+= _abs(Set[0].dis[S[j].u] - Set[0].dis[S[j].v]); ra
= Set[1].find(S[j].u), rb = Set[1].find(S[j].v); if
(ra != rb) { ans[S[j].lab]
= -1; continue; } ans[S[j].lab]
+= _abs(Set[1].dis[S[j].u] - Set[1].dis[S[j].v]); } } for(i
= 1; i <= Q; ++i) printf("%d\n",
ans[i]); return
0;}