Fermat point Figure -1400. Graph

2020-04-06 18:45:31

Problem Description:

There is a connected graph undirected acyclic, each edge by two vertices x [i], y [i ] will be described, the length of each side will be described by d [i].
Seeking such a point p, the other point such that pthe minimum distance and, if there are a plurality of such points p, the smallest number returned.

Sample

Example 1:

给出 x = `[1]`, y = `[2]`, d = `[3]`, 返回 `1`。
输入:
[1]
[2]
[3]
输出:
1

解释：
其他点到 1 的距离和为 3，其他点到 2 的距离和为 3，1 的编号较小。

Sample 2:

给出 x = `[1,2,2]`, y = `[2,3,4]`, d = `[1,1,1]`, 返回 `2`。
输入:
[1,2,2]
[2,3,4]
[1,1,1]
输出:
2

解释：
其他点到 1 的距离和为 5，其他点到 2 的距离和为 3，其他点到 3 的距离和为 5，其他点到 4 的距离和为 5。

Precautions

• 2 <= n, d[i] <= 10^5
• 1 <= x[i], y[i] <= n

Problem Solving:

Paths and trees and the like, but the value of the right side here.

Time complexity: O (n)

    Map<Integer, Set<int[]>> graph = new HashMap<>();
    Map<Integer, Integer> cnt = new HashMap<>();
    Map<Integer, Long> sum = new HashMap<>();
    List<long[]> record = new ArrayList<>();
    public int getFermatPoint(int[] x, int[] y, int[] d) {
        for (int i = 0; i < x.length; i++) {
            int from = x[i];
            int to = y[i];
            int w = d[i];
            if (!graph.containsKey(from)) graph.put(from, new HashSet<>());
            if (!graph.containsKey(to)) graph.put(to, new HashSet<>());
            graph.get(from).add(new int[]{to, w});
            graph.get(to).add(new int[]{from, w});
        }
        dfs1(x[0], -1);
        dfs2(x[0], -1);
        Collections.sort(record, (long[] o1, long[] o2) -> o1[0] == o2[0] ? Long.compare(o1[1], o2[1]) : Long.compare(o1[0], o2[0]));
        return (int)record.get(0)[1];
    }
    
    private void dfs2(int root, int parent) {
        record.add(new long[]{sum.get(root), root});
        for (int[] next : graph.get(root)) {
            if (next[0] == parent) continue;
            long w = (long)next[1];
            long curr = sum.get(root) + (long)(cnt.size() - cnt.get(next[0]) * 2) * w;
            sum.put(next[0], curr);
            dfs2(next[0], root);
        }
    }
    
    private void dfs1(int root, int parent) {
        int curr_cnt = 1;
        long curr_sum = 0;
        for (int[] next : graph.get(root)) {
            if (next[0] == parent) continue;
            dfs1(next[0], root);
            curr_cnt += cnt.get(next[0]);
            curr_sum += sum.get(next[0]) + (long)next[1] * (long)cnt.get(next[0]);
        }
        cnt.put(root, curr_cnt);
        sum.put(root, curr_sum);
    }