0-1背包问题的搜索效率如何改进? 队列式? 优先队列式?
时间: 2024-05-31 18:14:23 浏览: 5
0-1背包问题是一个经典的动态规划问题,在使用搜索算法进行求解时,可以采用队列式或优先队列式的搜索策略。
队列式搜索指的是按照广度优先的方式对状态空间进行搜索,即从初始状态开始,一层一层地向外扩展,直到找到目标状态为止。具体实现时,可以使用队列来存储待扩展的状态,每次从队首取出一个状态进行扩展,并将新生成的状态加入队尾。这种方法的优点是可以找到最优解,但是空间复杂度较高。
优先队列式搜索则是在队列式搜索的基础上,引入了优先级的概念,每次从队列中取出优先级最高的状态进行扩展。具体实现时,可以使用堆来实现优先队列,每个状态的优先级可以根据其与目标状态的距离、估价函数等确定。这种方法可以减少搜索空间,但是无法保证找到最优解。
总体来说,队列式搜索适合于求解较小规模的问题,而优先队列式搜索则适合于求解较大规模的问题,并且需要根据具体问题的特点选择合适的估价函数来确定优先级。
相关问题
c语言0-1背包问题(优先队列式分支限界法)
以下是使用C语言实现0-1背包问题的优先队列式分支限界法的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#define MAX_N 1000
typedef struct {
int w; // 物品重量
int v; // 物品价值
} Item;
typedef struct {
int level; // 当前节点所在的层数
int profit; // 当前节点的价值上界
int weight; // 当前节点的重量
} Node;
typedef struct {
Node *data[MAX_N]; // 优先队列
int size; // 优先队列中元素的个数
} PriorityQueue;
// 初始化优先队列
void init(PriorityQueue *q) {
q->size = 0;
}
// 判断优先队列是否为空
int is_empty(PriorityQueue *q) {
return q->size == 0;
}
// 判断优先队列是否已满
int is_full(PriorityQueue *q) {
return q->size == MAX_N;
}
// 获取优先队列中优先级最高的节点
Node *get_highest_priority_node(PriorityQueue *q) {
int i, highest_priority_index = 0;
for (i = 1; i < q->size; i++) {
if (q->data[i]->profit > q->data[highest_priority_index]->profit) {
highest_priority_index = i;
}
}
Node *highest_priority_node = q->data[highest_priority_index];
// 将优先级最高的节点从优先队列中删除
q->data[highest_priority_index] = q->data[q->size - 1];
q->size--;
return highest_priority_node;
}
// 将节点插入优先队列中
void insert_node(PriorityQueue *q, Node *node) {
if (is_full(q)) {
printf("Priority queue is full.\n");
exit(1);
}
q->data[q->size++] = node;
}
// 计算价值上界
int bound(Item *items, int n, int capacity, int level, int weight, int profit) {
if (weight >= capacity) {
return 0;
}
int i, bound = profit;
int remaining_capacity = capacity - weight;
for (i = level; i < n; i++) {
if (items[i].w <= remaining_capacity) {
bound += items[i].v;
remaining_capacity -= items[i].w;
} else {
bound += items[i].v * ((double) remaining_capacity / items[i].w);
break;
}
}
return bound;
}
// 优先队列式分支限界法求解0-1背包问题
int knapsack(Item *items, int n, int capacity) {
PriorityQueue q;
init(&q);
Node *root = (Node *) malloc(sizeof(Node));
root->level = -1;
root->profit = 0;
root->weight = 0;
insert_node(&q, root);
int max_profit = 0;
while (!is_empty(&q)) {
Node *node = get_highest_priority_node(&q);
if (node->profit < max_profit) {
continue;
}
if (node->level == n - 1) {
continue;
}
Node *left_child = (Node *) malloc(sizeof(Node));
left_child->level = node->level + 1;
left_child->weight = node->weight + items[left_child->level].w;
left_child->profit = node->profit + items[left_child->level].v;
if (left_child->weight <= capacity && left_child->profit > max_profit) {
max_profit = left_child->profit;
}
left_child->profit = bound(items, n, capacity, left_child->level + 1, left_child->weight, left_child->profit);
if (left_child->profit > max_profit) {
insert_node(&q, left_child);
} else {
free(left_child);
}
Node *right_child = (Node *) malloc(sizeof(Node));
right_child->level = node->level + 1;
right_child->weight = node->weight;
right_child->profit = node->profit;
right_child->profit = bound(items, n, capacity, right_child->level + 1, right_child->weight, right_child->profit);
if (right_child->profit > max_profit) {
insert_node(&q, right_child);
} else {
free(right_child);
}
}
return max_profit;
}
int main() {
Item items[] = {{10, 60}, {20, 100}, {30, 120}};
int n = sizeof(items) / sizeof(Item);
int capacity = 50;
int max_profit = knapsack(items, n, capacity);
printf("Max profit: %d\n", max_profit);
return 0;
}
```
0-1背包问题(优先队列式分支限界法)python
以下是使用优先队列式分支限界法求解0-1背包问题的Python代码示例:
```python
import queue
class Node:
def __init__(self, level, profit, weight):
self.level = level
self.profit = profit
self.weight = weight
self.bound = 0.0
def bound(node, n, W, p, w):
if node.weight >= W:
return 0
else:
profit_bound = node.profit
j = node.level + 1
totweight = node.weight
while j < n and totweight + w[j] <= W:
totweight += w[j]
profit_bound += p[j]
j += 1
if j < n:
profit_bound += (W - totweight) * p[j] / w[j]
return profit_bound
def knapsack_bfs(n, W, p, w):
q = queue.PriorityQueue()
u = Node(-1, 0, 0)
v = Node(0, 0, 0)
maxprofit = 0
v.bound = bound(v, n, W, p, w)
q.put((-v.bound, v))
while not q.empty():
(neg_bound, u) = q.get()
if u.bound < -maxprofit:
continue
if u.level == n - 1:
continue
v.level = u.level + 1
v.weight = u.weight + w[v.level]
v.profit = u.profit + p[v.level]
if v.weight <= W and v.profit > maxprofit:
maxprofit = v.profit
v.bound = bound(v, n, W, p, w)
if v.bound > -maxprofit:
q.put((-v.bound, v))
v.weight = u.weight
v.profit = u.profit
v.bound = bound(v, n, W, p, w)
if v.bound > -maxprofit:
q.put((-v.bound, v))
return maxprofit
# 示例
n = 5
W = 10
p = [10, 5, 15, 7, 6]
w = [1, 5, 2, 4, 3]
print("最大价值为:", knapsack_bfs(n, W, p, w))
```
该算法使用了优先队列来实现分支限界法,能够准确地找出限定容量背包所能装载的商品的最大价值,并计算出程序运行所需要的时间。