标题效果：自脑补。
思维：与维护两个维度和可设置为检查右。
注意，标题给予一堆关系的。我们应该加入两对关系。
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;
}
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本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/4746996.html，如需转载请自行联系原作者
