Colorful Tree

Colorful Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3373    Accepted Submission(s): 1461

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.

 

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.

 

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

2017 Multi-University Training Contest - Team 1

#pragma GCC optimize(2)

#include <bits/stdc++.h>

#define lowbit(x) x&(-x)
typedef long long ll;
using namespace std;
const int maxn = 2e5 + 100000;
ll n;
ll siz[maxn], sum[maxn], cnt;
ll color[maxn], vis[maxn];
struct node {
    int to, nx;
} o[maxn << 1];
int head[maxn];
ll res;

inline void add_edge(int u, int v) {
    o[++cnt] = (node) {v, head[u]};
    head[u] = cnt;
}

inline void solve(int cur, int fa) {
    siz[cur] = 1;
    ll pre = sum[color[cur]];
    ll e = 0;
    for (register int i = head[cur]; i; i = o[i].nx) {
        int to = o[i].to;
        if (to == fa)continue;
        solve(to, cur);
        siz[cur] += siz[to];
        ll nx = sum[color[cur]] - pre;
        ll fur = nx;
        nx = siz[to] - nx;
        res -= nx * (nx - 1) / 2;
        pre = sum[color[cur]];
        e += fur;
    }
    sum[color[cur]] += siz[cur] - e;
}

int Case;

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    while (scanf("%lld", &n) != EOF) {
        ll tot = 0;
        cnt = 0;
        memset(head, 0, sizeof(head));
        memset(vis, 0, sizeof(vis));
        memset(sum,0,sizeof(sum));
        for (register int i = 1; i <= n; ++i) {
            scanf("%d", &color[i]);
            if (!vis[color[i]]) {
                ++tot;
                vis[color[i]] = 1;
            }
        }
        for (register int i = 1, u, v; i < n; ++i) {
            scanf("%d%d", &u, &v);
            add_edge(u, v);
            add_edge(v, u);
        }
        res = tot * n * (n - 1) / 2;
        solve(1, -1);
        for (register int i = 1; i <= n; ++i) {
            if (!sum[i])continue;
            ll leave = n - sum[i];
            res -= leave * (leave - 1) / 2;
        }
        printf("Case #%d: ", ++Case);
        printf("%lld
", res);
    }
    return 0;
}