[IOI2011]Race

2020-03-17
IOI

题意

求树上路径长度为$K$的最小的边数,$n\leq 2\cdot 10^5$

题解

还是点分治,这次不能容斥了,对于一个rt计算的时候把每棵子树染色

更新答案的时候,如果当前$dis_l+dis_r=K$,不能直接l++,r—,要把其他能和l或者r匹配的也遍历一下

理论上来说来一个菊花图复杂度就直接上$\Theta(n^2)$,但是这样写开O2可过

更正确一点的写法是记录之前的其他子树对于每个dis的dep最小值,由于只要顾及和为$K$的情况,当前子树里的一个点只要与相加为$K​$的dis的dep最小值匹配即可,复杂度是可以保证的

然而我懒人一个,过了就算了

调试记录

没有遍历全(见上)

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
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
struct E{
int to, nxt, l;
}e[maxn << 1];
int head[maxn], tot = 0;
void addedge(int u, int v, int l){
e[++tot].to = v, e[tot].nxt = head[u], e[tot].l = l;
head[u] = tot;
}
bool vis[maxn];
int sz[maxn], Max[maxn], rt;
void getRt(int cur, int fa, int totsz){
sz[cur] = 1; Max[cur] = -1;
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (v == fa || vis[v]) continue;
getRt(v, cur, totsz);
sz[cur] += sz[v];
Max[cur] = max(Max[cur], sz[v]);
} Max[cur] = max(Max[cur], totsz - sz[cur]);
if (Max[cur] < Max[rt]) rt = cur;
}
struct A{
int dis, dep, c;
bool operator<(A x)const{
return dis < x.dis;
}
}; vector <A> v;
int dis[maxn], dep[maxn];
void getDis(int cur, int fa, int c){
v.push_back((A){dis[cur], dep[cur], c});
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (v == fa || vis[v]) continue;
dep[v] = dep[cur] + 1;
dis[v] = dis[cur] + e[i].l;
getDis(v, cur, c);
}
}
int n, K, ans = INF;
void calc(int cur){
int cntC = 0; v.clear();
v.push_back((A){0, 0, 0});
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (vis[v]) continue;
dep[v] = 1; dis[v] = e[i].l;
getDis(v, cur, ++cntC);
}
// for (int i = 0; i < v.size(); i++)
// for (int j = i + 1; j < v.size(); j++)
// if (v[i].c != v[j].c && v[i].dis + v[j].dis == K)
// ans = min(ans, v[i].dep + v[j].dep);
sort(v.begin(), v.end());
vector <A>::iterator l = v.begin(), r = v.end() - 1;
while (l < r){
if (l->dis + r->dis < K) ++l;
else if (l->dis + r->dis > K) --r;
else{
if (l->c != r->c) ans = min(ans, l->dep + r->dep);
vector <A>::iterator t = l + 1;
while (t->dis == l->dis){
if (t->c != r->c) ans = min(ans, t->dep + r->dep);
++t;
}
t = r - 1;
while (t->dis == r->dis){
if (t->c != l->c) ans = min(ans, t->dep + l->dep);
--t;
}
++l, --r;
}
}
}
void dfs(int cur, int totsz){
vis[cur] = 1;
calc(cur);
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (vis[v]) continue;
rt = 0; int ssz = sz[v] > sz[cur] ? totsz - sz[cur] : sz[v];
getRt(v, cur, ssz);
dfs(rt, ssz);
}
}
int main(){
// freopen("2.in", "r", stdin);
scanf("%d%d", &n, &K);
for (int u, v, l, i = 1; i < n; i++){
scanf("%d%d%d", &u, &v, &l); ++u, ++v;
addedge(u, v, l); addedge(v, u, l);
} rt = 0; Max[0] = INF;
getRt(1, 0, n);
dfs(rt, n);
if (ans != INF) printf("%d\n", ans);
else puts("-1");
return 0;
}