Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P1967 [NOIP2013 提高组] 货车运输

P1967 [NOIP2013 提高组] 货车运输

一辆车的最大运输的重量取决于路径上的限重的最小值,我们先求最小生成树,同时也是最小瓶颈树,保证答案不变。题目不保证所有点连通,所以这是一个森林。同样的,用树链剖分维护路径上的最小值。

查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e4 + 10, M = 2e4 + 10, inf = 1e5;
int n, m, q, d[N], p[N], f[N], w[N], rt[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
int stmp, sz[N], son[N], dfn[N], top[N], rnk[N];
struct Edge
{
int u, v, w;
bool operator<(Edge &_)
{
return w > _.w;
}
} e[M];
int find(int x)
{
return x == f[x] ? x : f[x] = find(f[x]);
}
struct Node
{
int l, r;
int mn;
} tr[N << 2];
void pushup(int x)
{
tr[x].mn = min(tr[x << 1].mn, tr[x << 1 | 1].mn);
}
void build(int x, int l, int r)
{
tr[x].l = l;
tr[x].r = r;
if (l == r)
{
tr[x].mn = w[rnk[l]];
return;
}
int mid = l + r >> 1;
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
pushup(x);
}
int query(int x, int l, int r)
{
if (tr[x].l >= l && tr[x].r <= r)
return tr[x].mn;
int res = inf;
int mid = tr[x].l + tr[x].r >> 1;
if (l <= mid)
res = min(res, query(x << 1, l, r));
if (r > mid)
res = min(res, query(x << 1 | 1, l, r));
return res;
}
int queryPath(int x, int y)
{
int res = inf;
while (top[x] != top[y])
{
if (d[top[x]] < d[top[y]])
swap(x, y);
res = min(res, query(1, dfn[top[x]], dfn[x]));
x = p[top[x]];
}
if (x == y)
return res;
if (d[x] > d[y])
swap(x, y);
return min(res, query(1, dfn[x] + 1, dfn[y]));
}
void dfs1(int x)
{
sz[x] = 1;
son[x] = -1;
for (int i = hd[x]; ~i; i = nxt[i])
if (!d[edg[i]])
{
d[edg[i]] = d[x] + 1;
p[edg[i]] = x;
w[edg[i]] = wt[i];
dfs1(edg[i]);
sz[x] += sz[edg[i]];
if (son[x] == -1 || sz[edg[i]] > sz[son[x]])
son[x] = edg[i];
}
}
void dfs2(int x, int t)
{
dfn[x] = ++stmp;
rnk[stmp] = x;
top[x] = t;
if (son[x] == -1)
return;
dfs2(son[x], t);
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != p[x] && edg[i] != son[x])
dfs2(edg[i], edg[i]);
}
void add(int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1; i <= n; i++)
f[i] = i;
for (int i = 0; i < m; i++)
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
sort(e, e + m);
for (int i = 0; i < m; i++)
if (find(e[i].u) != find(e[i].v))
{
f[f[e[i].u]] = f[e[i].v];
add(e[i].u, e[i].v, e[i].w);
add(e[i].v, e[i].u, e[i].w);
}
for (int i = 1; i <= n; i++)
if (!dfn[i])
{
d[i] = 1;
dfs1(i);
dfs2(i, i);
}
build(1, 1, n);
scanf("%d", &q);
for (int x, y; q; q--)
{
scanf("%d%d", &x, &y);
if (find(x) != find(y))
puts("-1");
else
printf("%d\n", queryPath(x, y));
}
return 0;
}