地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=6035
题面：
Colorful Tree
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1651 Accepted Submission(s): 675
Problem Description
There is a tree with n nodes, each of which has a type of color represented by an integer, where the color of node i is ci.
The path between each two different nodes is unique, of which we define the value as the number of different colors appearing in it.
Calculate the sum of values of all paths on the tree that has n(n−1)2 paths in total.
The path between each two different nodes is unique, of which we define the value as the number of different colors appearing in it.
Calculate the sum of values of all paths on the tree that has n(n−1)2 paths in total.
Input
The input contains multiple test cases.
For each test case, the first line contains one positive integers n, indicating the number of node. (2≤n≤200000)
Next line contains n integers where the i-th integer represents ci, the color of node i. (1≤ci≤n)
Each of the next n−1 lines contains two positive integers x,y (1≤x,y≤n,x≠y), meaning an edge between node x and node y.
It is guaranteed that these edges form a tree.
For each test case, the first line contains one positive integers n, indicating the number of node. (2≤n≤200000)
Next line contains n integers where the i-th integer represents ci, the color of node i. (1≤ci≤n)
Each of the next n−1 lines contains two positive integers x,y (1≤x,y≤n,x≠y), meaning an edge between node x and node y.
It is guaranteed that these edges form a tree.
Output
For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.
Sample Input
3
1 2 1
1 2
2 3
6
1 2 1 3 2 1
1 2
1 3
2 4
2 5
3 6
Sample Output
Case #1: 6
Case #2: 29
Source
思路：很巧妙的树形dp！
具体看这个博客把，我也是看了这个才懂的http://blog.csdn.net/Bahuia/article/details/76141574
#include <bits/stdc++.h> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef pair<int,int> PII; const double eps=1e-7; const double pi=acos(-1.0); const int K=1e6+7; const int mod=1e9+7; vector<int>mp[K]; LL n,cnt,ans,sz[K],col[K],vis[K],sum[K]; void dfs(int x,int f) { LL tmp,add,all=0; sz[x]=1; for(auto v:mp[x]) if(v!=f) { tmp=sum[col[x]]; dfs(v,x); sz[x]+=sz[v]; add=sum[col[x]]-tmp; ans+=(sz[v]-add)*(sz[v]-add-1)/2; all+=sz[v]-add; } sum[col[x]]+=all+1; } int main(void) { int cs=1; while(~scanf("%lld",&n)) { cnt=0,ans=0; memset(mp,0,sizeof mp); memset(sum,0,sizeof sum); memset(sz,0,sizeof sz); for(int i=1;i<=n;i++) scanf("%d",col+i),vis[col[i]]=1; for(int i=1,x,y;i<n;i++) scanf("%d%d",&x,&y),mp[x].PB(y),mp[y].PB(x); dfs(1,0); for(int i=1;i<=n;i++) if(vis[i]) ans+=(n-sum[i]-1)*(n-sum[i])/2,vis[i]=0,cnt++; printf("Case #%d: %lld ",cs++,cnt*n*(n-1LL)/2LL-ans); } return 0; }