树形dp基础题 求带点权树直径
然后这个点权就是每个点的点度(son[])...
所以可以简化一下
按照正常的套路维护从根开始的最长链和次长链
dp数组存的时候+1改成+son[x]-1就可以了
所以dp[x] = maxx + maxxx + son[x] - 1
(maxx 和 maxxx 维护最长链和次长链)
同时ans取max
TIme cost : 45min
Code:
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 #include<cmath>
5 #include<queue>
6 #define ms(a,b) memset(a,b,sizeof a)
7 #define rep(i,a,n) for(int i = a;i <= n;i++)
8 #define per(i,n,a) for(int i = n;i >= a;i--)
9 #define inf 2147483647
10 using namespace std;
11 typedef long long ll;
12 typedef double D;
13 #define eps 1e-8
14 ll read() {
15 ll as = 0,fu = 1;
16 char c = getchar();
17 while(c < '0' || c > '9') {
18 if(c == '-') fu = -1;
19 c = getchar();
20 }
21 while(c >= '0' && c <= '9') {
22 as = as * 10 + c - '0';
23 c = getchar();
24 }
25 return as * fu;
26 }
27 //head
28 const int N = 300003;
29 int n;
30 int head[N],nxt[N<<1],mo[N<<1],cnt;
31 int son[N];
32 void _add(int x,int y) {
33 son[x]++;
34 mo[++cnt] = y;
35 nxt[cnt] = head[x];
36 head[x] = cnt;
37 }
38 void add(int x,int y) {
39 if(x^y)_add(x,y),_add(y,x);
40 }
41
42 int dp[N],ans;
43 void dfs(int x,int f) {
44 int maxx = 0,maxxx = 0;
45 for(int i = head[x];i;i = nxt[i]) {
46 int sn = mo[i];
47 if(sn == f) continue;
48 dfs(sn,x);
49 if(dp[sn] > maxx) maxxx = maxx,maxx = dp[sn];
50 else if(dp[sn] > maxxx) maxxx = dp[sn];
51 dp[x] = max(dp[x],dp[sn] + son[x] - 1);
52 }
53 ans = max(ans,maxx + maxxx + son[x] - 1);
54 }
55
56 int main() {
57 n = read(),read();
58 rep(i,2,n) add(read(),read());
59 rep(i,1,n) dp[i] = 1;
60 dfs(1,0),printf("%d
",ans);
61 return 0;
62 }