志愿者调配

/*
 * 思路：
 * 1. 逆向操作所有的调配方案，恢复初始状态。
 * 2. 建立场馆和其相邻场馆的图。
 * 3. 根据最终状态和总人数计算初始状态。
 */

import java.util.*;
import java.util.stream.*;

public class VolunteerAllocationStream {
    public int[] findInitialVolunteers(int[] finalCnt, int totalNum, int[][] edges, int[][] plans) {
        int n = finalCnt.length + 1;
        int[] initialCnt = new int[n];
        initialCnt[0] = totalNum - Arrays.stream(finalCnt).sum();
        System.arraycopy(finalCnt, 0, initialCnt, 1, finalCnt.length);
        Map<Integer, List<Integer>> graph = Arrays.stream(edges).collect(Collectors.groupingBy(edge -> edge[0], Collectors.mapping(edge -> edge[1], Collectors.toList())));
        Arrays.stream(edges).forEach(edge -> graph.computeIfAbsent(edge[1], k -> new ArrayList<>()).add(edge[0]));
        List<int[]> reversedPlans = Arrays.stream(plans).collect(Collectors.toList());
        Collections.reverse(reversedPlans);
        reversedPlans.forEach(plan -> {
            int type = plan[0];
            int idx = plan[1];
            if (type == 1) {
                initialCnt[idx] *= 2;
            } else if (type == 2) {
                graph.getOrDefault(idx, new ArrayList<>()).forEach(neighbor -> initialCnt[neighbor] -= initialCnt[idx]);
            } else if (type == 3) {
                graph.getOrDefault(idx, new ArrayList<>()).forEach(neighbor -> initialCnt[neighbor] += initialCnt[idx]);
            }
        });
        return initialCnt;
    }
}