0-1背包问题的动态规划求解

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思路：

设value_get[i][j]为把前i个物品装入容量为j的背包所能获得的最大价值

需考虑能否装入物品i，以及是否装入物品i，递归式为：

value_get[i][j] = max{ value_get[i-1][j], value_get[i-1][ j-weight[i] ] + value[i] }

c语言代码如下：

/*************************************************************************    > File Name: knapsack_dynamic.c    > Author: arui    > Mail: 450252722@qq.com     > Created Time: Wed 26 Feb 2014 01:43:14 PM CST ************************************************************************/#include <stdio.h>#include <stdlib.h>int knapsack_dynamic(int capacity,            int *weight,int *value,int *choice, int obj_num){int ** value_get = (int**)malloc(sizeof(int*) * (obj_num + 1));for(int i = 0; i <= obj_num; ++i){value_get[i] = (int*)malloc(sizeof(int) * (capacity + 1));}//initializefor(int j = 0; j <= capacity; ++j){value_get[0][j] = 0;}for(int i = 0; i <= obj_num; ++i){value_get[i][0] = 0;}for(int i=1; i <= obj_num; ++i){for(int j=1; j <= capacity; ++j){value_get[i][j] = value_get[i-1][j];if(weight[i] <= j){int temp_value = value_get[i-1][j-weight[i]] + value[i];if(temp_value > value_get[i][j])value_get[i][j] = temp_value;}}}//construct the choiceint j = capacity;for(int i = obj_num; i > 0; --i){if(value_get[i][j] > value_get[i-1][j]){choice[i] = 1;j = j - weight[i];}}//free the space for(int i = 0; i <= obj_num; ++i){free(value_get[i]);}free(value_get);return value_get[obj_num][capacity];}int main(int argc, char** argv){int *weight;int *value;int *choice;int capacity;int obj_num;freopen("test.dat","r", stdin);while(scanf("%d%d",&capacity, &obj_num) != EOF){weight = (int*)malloc((obj_num + 1) * sizeof(int));value = (int*)malloc((obj_num + 1) * sizeof(int));choice = (int*)malloc((obj_num + 1) * sizeof(int));for(int i = 1; i <= obj_num; ++i){scanf("%d", weight + i);}for(int i = 1; i <= obj_num; ++i){scanf("%d", value + i);}int max_value = knapsack_dynamic(capacity,weight, value, choice, obj_num);printf("max_value = %d\n", max_value);for(int i = 1; i <= obj_num; ++i){printf("%d", choice[i]);}printf("\n");//free the spacefree(weight);free(value);free(choice);}return 0;}

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