[AGC004F]Namori

2020-02-08
Atcoder

题意

给出一棵树或基环树,开始每个节点都是白色

每次可以选择相连的颜色相同的节点反转颜色,求最小次数,使得节点全黑

题解

先考虑树,分深度奇偶考虑,等价地认为深度奇数上有一个球,偶数上有一个洞

令那么$s_u$为子树中球/洞的个数,那么$ans=\sum|s_i|$,若根节点s不为0则无解

对于基环树,把一条边放到最后考虑,顺序没关系,随便断开一条边

对于奇环,两个点深度奇偶相同,一起变颜色可以多出两个球/洞

对于偶环,两个点深度就不同,深度小的一个点$(设为cv)$可以把球给另外一个点,绕一圈再回来还给它,那么环上的点(cv除外)可以加或减同一个数,要使$\sum |s_i-k|$最小,显然k是这些数的中位数时最优

调试记录

sf变量混淆

(dep[cu]-dep[cv])%2没加括号

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
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
struct E{
int to, nxt;
}e[maxn << 1];
int head[maxn], tot = 0;
void addedge(int u, int v){
e[++tot].to = v, e[tot].nxt = head[u];
head[u] = tot;
}
int s[maxn], dep[maxn], f[maxn]; int cu, cv;
void dfs(int cur, int fa){
dep[cur] = dep[fa] + 1;
f[cur] = fa;
if (dep[cur] & 1) s[cur] = 1;
else s[cur] = -1;
for (int i = head[cur]; i; i = e[i].nxt){
int v = e[i].to;
if (v != fa){
if (dep[v]){
if (cu == 0) cu = cur, cv = v;
}
else dfs(v, cur), s[cur] += s[v];
}
}
}
int n, m;
int abs(int x){return x < 0 ? -x : x;}
void work1(){
if (s[1] != 0){puts("-1"); exit(0);}
int ans = 0;
for (int i = 1; i <= n; i++) ans += abs(s[i]);
printf("%d\n", ans); exit(0);
}
void work2(){
if (s[1] % 2 != 0){puts("-1"); exit(0);}
int t = -s[1] / 2;
int ans = abs(t);
for (; cu; cu = f[cu]) s[cu] += t;
for (; cv; cv = f[cv]) s[cv] += t;
for (int i = 1; i <= n; i++) ans += abs(s[i]);
printf("%d\n", ans); exit(0);
}
vector <int> v;
void work3(){
if (s[1] != 0){puts("-1"); exit(0);}
for (; cu != cv; cu = f[cu]) dep[cu] = 0, v.push_back(s[cu]);
sort(v.begin(), v.end());
int mid = v[(v.size() - 1) / 2];
int ans = abs(mid);
for (int i = 1; i <= n; i++)
if (dep[i] != 0) ans += abs(s[i]);
else ans += abs(s[i] - mid);
printf("%d\n", ans); exit(0);
}
int main(){
scanf("%d%d", &n, &m);
for (int u, v, i = 1; i <= m; i++){
scanf("%d%d", &u, &v);
addedge(u, v); addedge(v, u);
} dfs(1, 0);
if (m == n - 1) work1();
if ((dep[cu] - dep[cv]) % 2 == 0) work2();
work3();
return 0;
}