init
parent
c187397c3e
commit
b91c2d09a4
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@ -21,7 +21,6 @@ import org.springframework.util.CollectionUtils;
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import java.math.BigDecimal;
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import java.util.*;
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import java.util.function.Consumer;
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import java.util.stream.Collectors;
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import static org.qinan.safetyeval.domain.exception.ErrorCode.PERSON_NOT_ENOUGH;
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@ -385,13 +384,11 @@ public class ComplianceCheckUtil {
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}
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// 数据来源：industryProfessionalPersonCos（备案人员自带专业能力）+ certCapabilityList（证书专业能力）
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// 4. 贪心分配（一人一岗）：按可选人数从少到多处理，优先满足最紧缺的专业能力
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Map<String, Integer> assignedMap = assignPersonnelGreedy(personCapabilityMap, requiredMap);
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boolean allMet = canFullyAssign(personCapabilityMap, requiredMap);
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// 5. 判断是否所有专业能力都达标，并生成失败消息
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boolean allMet = true;
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IndustryEnum industryEnum = IndustryEnum.ofCode(standardsDOList.get(0).getIndustryCode());
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String industryName = industryEnum != null ? industryEnum.getValue() : standardsDOList.get(0).getIndustryCode();
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@ -417,53 +414,94 @@ public class ComplianceCheckUtil {
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return allMet ? Pair.of(true, null) : Pair.of(false, failureMsg.toString());
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}
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/**
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* 一人一岗贪心分配
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* <p>
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* 每个人员最多分配到一个专业能力岗位，人员不能重复使用。
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* 按可选人员池从小到大排序优先处理最紧缺的专业能力，尽量满足所有要求。
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* <p>
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* 注意：贪心为启发式，不保证全局最优，但满足"只需判断能否达标"的场景。
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* 判断是否存在一种分配方案，使所有能力都达到要求人数
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*
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* @param personCapabilityMap 人员→可担任专业能力code集合（同一人可能具备多个专业能力）
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* @param requiredMap 专业能力code→要求人数
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* @return 专业能力code→实际分配人数
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* @param personCapabilityMap 人员ID → 具备的能力code集合
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* @param requiredMap 能力code → 需求人数
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* @return true 当且仅当存在满足所有需求的分配
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*/
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private static Map<String, Integer> assignPersonnelGreedy(Map<Long, Set<String>> personCapabilityMap
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, Map<String, Long> requiredMap) {
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// 构建 专业能力code → 可用人员集合
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Map<String, Set<Long>> capabilityPersonMap = new HashMap<>();
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public static boolean canFullyAssign(
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Map<Long, Set<String>> personCapabilityMap, Map<String, Long> requiredMap) {
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// ---- 1. 构图 ----
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// 节点编号分配
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// 源点: 0
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// 人员节点: 1 .. P
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// 能力节点: P+1 .. P+C
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// 汇点: P+C+1
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List<Long> personIds = new ArrayList<>(personCapabilityMap.keySet());
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List<String> capCodes = new ArrayList<>(requiredMap.keySet());
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int P = personIds.size();
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int C = capCodes.size();
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int S = 0;
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int T = P + C + 1;
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int totalNodes = T + 1;
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Dinic dinic = new Dinic(totalNodes);
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// 建立人员ID到节点编号的映射
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Map<Long, Integer> personNode = new HashMap<>();
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for (int i = 0; i < P; i++) {
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personNode.put(personIds.get(i), i + 1);
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}
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// 建立能力code到节点编号的映射
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Map<String, Integer> capNode = new HashMap<>();
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for (int i = 0; i < C; i++) {
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capNode.put(capCodes.get(i), P + 1 + i);
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}
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long totalRequired = 0;
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// 源点 → 人员 (容量1)
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for (int i = 0; i < P; i++) {
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dinic.addEdge(S, personNode.get(personIds.get(i)), 1);
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}
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// 人员 → 能力 (容量1)
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for (Map.Entry<Long, Set<String>> entry : personCapabilityMap.entrySet()) {
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Long pid = entry.getKey();
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int u = personNode.get(pid);
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for (String cap : entry.getValue()) {
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if (requiredMap.containsKey(cap)) {
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capabilityPersonMap.computeIfAbsent(cap, k -> new HashSet<>()).add(pid);
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if (capNode.containsKey(cap)) { // 仅连有需求的能力
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dinic.addEdge(u, capNode.get(cap), 1);
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}
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}
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}
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// 按可选人数从少到多排序，优先满足最紧缺的专业能力
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List<String> capOrder = new ArrayList<>(requiredMap.keySet());
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capOrder.sort(Comparator.comparingInt(c -> capabilityPersonMap.getOrDefault(c, Collections.emptySet()).size()));
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// 能力 → 汇点 (容量 = 需求人数)
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for (Map.Entry<String, Long> entry : requiredMap.entrySet()) {
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String cap = entry.getKey();
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long req = entry.getValue();
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totalRequired += req;
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int v = capNode.get(cap);
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dinic.addEdge(v, T, req);
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}
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Set<Long> usedPersonIds = new HashSet<>();
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Map<String, Integer> assignedMap = new LinkedHashMap<>();
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for (String capCode : capOrder) {
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long required = requiredMap.get(capCode);
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Set<Long> candidates = capabilityPersonMap.get(capCode);
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int assigned = 0;
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if (candidates != null) {
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for (Long pid : candidates) {
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if (assigned >= required) {
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break;
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// ---- 2. 计算最大流 ----
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long maxFlow = dinic.maxFlow(S, T);
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return maxFlow == totalRequired;
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}
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if (usedPersonIds.add(pid)) {
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assigned++;
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}
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}
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}
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assignedMap.put(capCode, assigned);
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}
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return assignedMap;
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// 简单测试
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public static void main(String[] args) {
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Map<Long, Set<String>> personMap = new HashMap<>();
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personMap.put(1L, new HashSet<>(Arrays.asList("A", "B")));
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personMap.put(2L, new HashSet<>(Arrays.asList("A", "B")));
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personMap.put(3L, new HashSet<>(Arrays.asList("A")));
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personMap.put(4L, new HashSet<>(Arrays.asList("B")));
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personMap.put(5L, new HashSet<>(Arrays.asList("A")));
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Map<String, Long> required = new HashMap<>();
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required.put("A", 2L);
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required.put("B", 2L);
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System.out.println(canFullyAssign(personMap, required)); // true
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// 让需求不可满足：A需要3人，B需要2人
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required.put("A", 3L);
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System.out.println(canFullyAssign(personMap, required)); // false
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}
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}
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@ -0,0 +1,82 @@
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package org.qinan.safetyeval.infrastructure.utils;
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import java.util.*;
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/**
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* Dinic 算法实现
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*/
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public class Dinic {
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static class Edge {
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int to, rev;
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long cap;
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Edge(int to, int rev, long cap) {
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this.to = to;
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this.rev = rev;
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this.cap = cap;
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}
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}
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List<Edge>[] graph;
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int[] level, iter;
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@SuppressWarnings("unchecked")
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Dinic(int n) {
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graph = new ArrayList[n];
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for (int i = 0; i < n; i++) graph[i] = new ArrayList<>();
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level = new int[n];
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iter = new int[n];
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}
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void addEdge(int from, int to, long cap) {
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graph[from].add(new Edge(to, graph[to].size(), cap));
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graph[to].add(new Edge(from, graph[from].size() - 1, 0));
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}
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void bfs(int s) {
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Arrays.fill(level, -1);
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Queue<Integer> q = new LinkedList<>();
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level[s] = 0;
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q.offer(s);
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while (!q.isEmpty()) {
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int v = q.poll();
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for (Edge e : graph[v]) {
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if (e.cap > 0 && level[e.to] < 0) {
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level[e.to] = level[v] + 1;
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q.offer(e.to);
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}
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}
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}
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}
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long dfs(int v, int t, long f) {
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if (v == t) return f;
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for (int i = iter[v]; i < graph[v].size(); i++) {
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iter[v] = i;
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Edge e = graph[v].get(i);
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if (e.cap > 0 && level[v] < level[e.to]) {
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long d = dfs(e.to, t, Math.min(f, e.cap));
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if (d > 0) {
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e.cap -= d;
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graph[e.to].get(e.rev).cap += d;
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return d;
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}
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}
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}
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return 0;
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}
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long maxFlow(int s, int t) {
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long flow = 0;
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while (true) {
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bfs(s);
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if (level[t] < 0) break;
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Arrays.fill(iter, 0);
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long f;
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while ((f = dfs(s, t, Long.MAX_VALUE)) > 0) {
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flow += f;
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}
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}
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return flow;
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}
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}
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