add

2026-06-15 09:00:38 +08:00
parent fec66377d5
commit 4640c5e02b
191 changed files with 6046 additions and 0 deletions
@@ -0,0 +1,176 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+#define N 3            // 物品数量
+#define CAPACITY 30    // 背包容量
+#define MAX_NODES (N * 2) // 估计的最大节点数
+
+// 物品结构体
+typedef struct {
+    int weight;
+    int value;
+} Item;
+
+// 活结点结构体
+typedef struct {
+    int level;  // 当前节点的层级
+    int profit; // 当前节点的价值
+    int weight; // 当前节点的重量
+    float bound; // 当前节点的上界
+} Node;
+
+// 优先队列（最大堆）
+typedef struct {
+    Node nodes[MAX_NODES];
+    int size;
+} PriorityQueue;
+
+// 初始化优先队列
+void initQueue(PriorityQueue* pq) {
+    pq->size = 0;
+}
+
+// 插入节点到优先队列
+void insert(PriorityQueue* pq, Node node) {
+    if (pq->size == MAX_NODES) {
+        printf("Priority queue is full.\n");
+        return;
+    }
+    int i = pq->size++;
+    while (i > 0 && pq->nodes[(i - 1) / 2].bound < node.bound) {
+        pq->nodes[i] = pq->nodes[(i - 1) / 2];
+        i = (i - 1) / 2;
+    }
+    pq->nodes[i] = node;
+}
+
+// 删除并返回优先队列的最大元素
+Node removeMax(PriorityQueue* pq) {
+    if (pq->size <= 0) {
+        printf("Priority queue is empty.\n");
+        exit(1);
+    }
+    Node max = pq->nodes[0];
+    pq->nodes[0] = pq->nodes[--pq->size];
+    int i = 0;
+    while (2 * i + 1 < pq->size) {
+        int left = 2 * i + 1;
+        int right = 2 * i + 2;
+        int largest = i;
+        if (left < pq->size && pq->nodes[left].bound > pq->nodes[largest].bound)
+            largest = left;
+        if (right < pq->size && pq->nodes[right].bound > pq->nodes[largest].bound)
+            largest = right;
+        if (largest != i) {
+            Node temp = pq->nodes[i];
+            pq->nodes[i] = pq->nodes[largest];
+            pq->nodes[largest] = temp;
+            i = largest;
+        } else break;
+    }
+    return max;
+}
+
+// 计算节点的上界（bound），这是通过贪心策略估计的最大可能价值
+float calculateBound(Node u, int n, int W, Item items[]) {
+    if (u.weight >= W) return 0;
+
+    int j = u.level + 1;
+    float bound = u.profit;
+    int totalWeight = u.weight;
+
+    // 从当前节点继续往下选择物品的贪心解（按单位价值排序）
+    while (j < n && totalWeight + items[j].weight <= W) {
+        totalWeight += items[j].weight;
+        bound += items[j].value;
+        j++;
+    }
+
+    if (j < n) {
+        bound += (W - totalWeight) * ((float)items[j].value / items[j].weight);
+    }
+
+    return bound;
+}
+
+// 比较函数，用于qsort排序
+int compareItems(const void* a, const void* b) {
+    Item* itemA = (Item*)a;
+    Item* itemB = (Item*)b;
+    float unitValueA = (float)itemA->value / itemA->weight;
+    float unitValueB = (float)itemB->value / itemB->weight;
+    return (unitValueB - unitValueA) > 0 ? 1 : -1;
+}
+
+// 0-1 背包问题的分支限界法
+int knapsackBranchAndBound(Item items[], int n, int W) {
+    // 按照单位价值对物品进行排序
+    qsort(items, n, sizeof(Item), compareItems);
+
+    PriorityQueue pq;
+    initQueue(&pq);
+
+    // 初始化第一个节点
+    Node u = { -1, 0, 0, 0.0 };
+    insert(&pq, u);
+
+    int maxProfit = 0;
+
+    while (pq.size > 0) {
+        // 取出优先队列中的最大元素
+        u = removeMax(&pq);
+
+        // 如果当前节点的 bound 值不大于最大收益，则剪枝
+        if (u.bound <= maxProfit) continue;
+
+        // 尝试将下一个物品放入背包
+        Node v = u;
+        v.level++;
+
+        if (v.level < n) {
+            // 选择第 v.level 个物品
+            if (v.weight + items[v.level].weight <= W) {
+                v.weight += items[v.level].weight;
+                v.profit += items[v.level].value;
+                if (v.profit > maxProfit) {
+                    maxProfit = v.profit;
+                }
+                v.bound = calculateBound(v, n, W, items);
+                if (v.bound > maxProfit) {
+                    insert(&pq, v);
+                }
+            }
+
+            // 不选择第 v.level 个物品
+            v = u;
+            v.level++;
+            v.bound = calculateBound(v, n, W, items);
+            if (v.bound > maxProfit) {
+                insert(&pq, v);
+            }
+        }
+    }
+
+    return maxProfit;
+}
+
+int main() {
+    // 已经填写的数据
+    int weightArr[N] = {16, 15, 15}; // 每个物品的重量
+    int valueArr[N] = {45, 25, 25};  // 每个物品的价值
+
+    // 初始化物品数组
+    Item items[N];
+    for (int i = 0; i < N; i++) {
+        items[i].weight = weightArr[i];
+        items[i].value = valueArr[i];
+    }
+
+    // 调用分支限界法求解0-1背包问题
+    int maxProfit = knapsackBranchAndBound(items, N, CAPACITY);
+
+    // 输出最大收益
+    printf("Maximum profit: %d\n", maxProfit);
+
+    return 0;
+}