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 40010
int
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 10010
struct
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 40010
int
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;
}