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樹形dp基礎題 求帶點權樹直徑

然後這個點權就是每個點的點度(son[])...

所以可以簡化一下

按照正常的套路維護從根開始的最長鏈和次長鏈

dp陣列存的時候+1改成+son[x]-1就可以了

所以dp[x] = maxx + maxxx + son[x] - 1

(maxx 和 maxxx 維護最長鏈和次長鏈)

同時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\n",ans);
61     return 0;
62 }