題目傳送門：http://acm.hdu.edu.cn/showproblem.php?pid=6035

題意：

給出一顆樹， 樹上路徑權值等於路徑上不同顏色的數量，求所有路徑權值之和；

解題思路：

大佬講的很清楚了：https://blog.csdn.net/qust1508060414/article/details/76360368

AC code：

 1 #include <bits/stdc++.h>
 2 #define INF 0x3f3f3f3f
 3 #define LL long long
 4 using namespace std;
 5
 
 const int MAXN = 2e5+10;
 6 int c[MAXN];
 7 bool vis[MAXN];
 8 vector<int>e[MAXN];
 9 LL sum[MAXN], Size[MAXN];
10 LL ans;
11 
12 void dfs(int x, int fa)
13 {
14     Size[x] = 1;
15     sum[c[x]]++;
16     LL pre = sum[c[x]];
17     int len = e[x].size();
18     for(int i = 0; i < len; i++){
 
19         if(e[x][i] == fa) continue;
20         dfs(e[x][i], x);
21         Size[x]+=Size[e[x][i]];
22         LL Count = Size[e[x][i]]-(sum[c[x]]-pre);
23         ans = ans+(Count*(Count-1))/2;
24         sum[c[x]]+=Count;
25         pre = sum[c[x]];
26     }
27 }
28 
29 int main()
30 {
 
31     int N, cas = 1;
32     LL num = 0, ct = 0;
33     while(~scanf("%d", &N)){
34         num = 0; ans = 0;
35         memset(sum, 0, sizeof(sum));
36         memset(vis, 0, sizeof(vis));
37         for(int i = 1; i <= N; i++){
38             e[i].clear();
39             scanf("%d", &c[i]);
40             if(!vis[c[i]]){
41                 vis[c[i]] = true;
42                 num++;
43             }
44         }
45         int u, v;
46         for(int i = 1; i < N; i++){
47             scanf("%d %d", &u, &v);
48             e[u].push_back(v);
49             e[v].push_back(u);
50         }
51         dfs(1, -1);
52         LL res = (LL)(1LL*N*(N-1))/2*num;
53         for(int i = 1; i <= N; i++){
54             if(vis[i]){
55                 ct = N-sum[i];
56                 ans+=ct*(ct-1)/2;
57             }
58         }
59         printf("Case #%d: %lld\n", cas++, res-ans);
60     }
61     return 0;
62 }

HDU 6015 Colorful Tree（2017多校）