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73 lines
2.0 KiB
73 lines
2.0 KiB
/**
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* File: unbounded_knapsack.c
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* Created Time: 2023-10-02
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* Author: Zuoxun (845242523@qq.com)
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*/
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#include "../utils/common.h"
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/* 求最大值 */
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int max(int a, int b) {
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return a > b ? a : b;
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}
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/* 完全背包：动态规划 */
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int unboundedKnapsackDP(int wgt[], int val[], int cap, int wgtSize) {
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int n = wgtSize;
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// 初始化 dp 表
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int dp[n + 1][cap + 1];
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memset(dp, 0, sizeof(dp));
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// 状态转移
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for (int i = 1; i <= n; i++) {
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for (int c = 1; c <= cap; c++) {
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if (wgt[i - 1] > c) {
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// 若超过背包容量，则不选物品 i
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dp[i][c] = dp[i - 1][c];
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} else {
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// 不选和选物品 i 这两种方案的较大值
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dp[i][c] = max(dp[i - 1][c], dp[i][c - wgt[i - 1]] + val[i - 1]);
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}
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}
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}
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return dp[n][cap];
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}
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/* 完全背包：空间优化后的动态规划 */
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int unboundedKnapsackDPComp(int wgt[], int val[], int cap, int wgtSize) {
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int n = wgtSize;
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// 初始化 dp 表
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int dp[cap + 1];
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memset(dp, 0, sizeof(dp));
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// 状态转移
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for (int i = 1; i <= n; i++) {
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for (int c = 1; c <= cap; c++) {
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if (wgt[i - 1] > c) {
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// 若超过背包容量，则不选物品 i
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dp[c] = dp[c];
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} else {
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// 不选和选物品 i 这两种方案的较大值
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dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1]);
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}
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}
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}
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return dp[cap];
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}
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/* Driver code */
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int main() {
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int wgt[] = {1, 2, 3};
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int val[] = {5, 11, 15};
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int wgtSize = sizeof(wgt) / sizeof(wgt[0]);
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int cap = 4;
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// 动态规划
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int res = unboundedKnapsackDP(wgt, val, cap, wgtSize);
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printf("不超过背包容量的最大物品价值为 %d\n", res);
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// 空间优化后的动态规划
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res = unboundedKnapsackDPComp(wgt, val, cap, wgtSize);
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printf("不超过背包容量的最大物品价值为 %d\n", res);
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return 0;
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}
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