问题描述
我已经为用 C 编写背包问题困惑了两个星期。
由于我没有设法让它直接与结构一起工作,所以我用一个额外的数组做了一个解决方案。如果您知道如何直接使用结构体,那就太好了!
但主要问题是,当我得到多个具有相同值的解时,我只想得到权重最小的解。
到目前为止我的代码:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <errno.h>
#include <stdbool.h>
#include <string.h>
#define TEAMS 5
#define max_players 20
typedef struct Team
{
char name [10];
int players;
bool Is_Playing;
int points;
} Team;
struct Team one = { "Ones",12,true,18 };
struct Team two = { "Twos",4,10 };
struct Team three = { "Threes",5,9 };
struct Team four = { "Fours",22 };
struct Team five = { "Fives",8,15 };
typedef struct selection_entry
{
int points;
int team;
struct selection_entry *prev;
} selection_entry;
Team *teams[TEAMS] = { &one,&two,&three,&four,&five };
int selection(int players[],const int *selection_points,size_t n,int capacity,int **solution)
{
int i,j;
selection_entry **table;
int result;
selection_entry *head;
/* Allocate the table */
table = malloc((n + 1) * sizeof(selection_entry *));
for (i = 0; i <= n; i++) {
table[i] = malloc((capacity + 1) * sizeof(selection_entry));
}
/* Calculate the points and build chains */
for (i = 0; i <= n; i++) {
for (j = 0; j <= capacity; j++) {
if (i == 0 || j == 0) {
/* Initialising the first row or column */
table[i][j].points = 0;
table[i][j].team = 0;
table[i][j].prev = NULL;
}
else if (players[i - 1] <= j) {
/* Can add team */
if (selection_points[i - 1] + table[i - 1][j - players[i - 1]].points
> table[i - 1][j].points) {
/* Add team */
table[i][j].points = selection_points[i - 1] + table[i - 1][j - players[i - 1]].points;
table[i][j].team = i - 1;
table[i][j].prev = &table[i - 1][j - players[i - 1]];
}
else {
/* Don't add team */
table[i][j].points = table[i - 1][j].points;
table[i][j].team = table[i - 1][j].team;
table[i][j].prev = table[i - 1][j].prev;
}
}
else {
/* Don't add team */
table[i][j].points = table[i - 1][j].points;
table[i][j].team = table[i - 1][j].team;
table[i][j].prev = table[i - 1][j].prev;
}
}
}
/* Read back the solution */
*solution = calloc(n,sizeof(int));
for (i = 0,head = &table[n][capacity];
head->prev != NULL;
head = head->prev,i++) {
(*solution)[head->team] = 1;
}
result = table[n][capacity].points;
for (i = 0; i <= n; i++) {
free(table[i]);
}
free(table);
return result;
}
int GetSelectionArraySize()
{
int s=0;
for (int i = 0; i < TEAMS; i++)
{
if (teams[i]->Is_Playing)
{
s++;
}
}
return s;
}
void main()
{
int a_size = GetSelectionArraySize();
int players[a_size];
int selection_points[a_size];
int i,j=0;
for (int i = 0; i < TEAMS; i++)
{
if (teams[i]->Is_Playing)
{
players[j] = teams[i]->players;
selection_points[j] = teams[i]->points;
j++;
}
}
const int capacity = max_players;
const size_t n = sizeof(players) / sizeof(players[0]);
int *solution;
int points = selection(players,selection_points,n,capacity,&solution);
fprintf(stdout,"Value: %d\n",points);
for (i = 0; i < n; i++)
{
if (solution[i])
{
fprintf(stdout,"Team %d with %d players and %d points.\n",i,players[i],selection_points[i]);
}
}
free(solution);
}
问题: 我无法弄清楚为什么这不能正常工作,以及我如何得到它来为我提供最佳解决方案(最高分,但在最大指定玩家数量内的玩家最少)。
子问题: 数组解决方法让我很恼火,但我无法直接使用原始结构数组...
在此先非常感谢您!!!
最诚挚的问候!
拉尔夫
解决方法
我不会详细介绍您的代码,而是让我们看看问题和类似背包的解决方案。
基本上,您的解决方案的相关属性是 value
和 weight
,其中 weight
受到约束并且解决方案 A 优于解决方案 B,如果 A.value > B.value
或如果A.value = B.value AND A.weight < B.weight
只要 A 和 B 满足约束。
因此,给定潜在项 [item1,[others...]]
的列表,其中包括 items1 的解决方案是可能的,您需要找到更好的解决方案
subresult1 = solve([others...],weightconstraint - item1.weight)
candidate1 = (subresult1.value + item1.value,subresult1.weight + item1.weight)
candidate2 = solve([others...],weightconstraint)
if (candidate1.value > candidate2.value) return candidate1
if (candidate1.value == candidate2.value && candidate1.weight < candidate2.weight) return candidate1
return candidate2
我希望这段伪代码足够清晰,可以解释要实现的逻辑。