Colorful Tree

时间:2019-09-03
本文章向大家介绍Colorful Tree，主要包括Colorful Tree使用实例、应用技巧、基本知识点总结和需要注意事项，具有一定的参考价值，需要的朋友可以参考一下。

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\n", res);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/czy-power/p/11451657.html

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