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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1114
思路分析:该问题要求为多重背包问题,使用多重背包的解法即可;假设dp[v]表示容量为v的背包中能够装下的最少的价值,因为一件物品可以装无限数次,所以可以得到递推公式: dp[v] = Min(dp[v], dp[v- c[i]] + w[i]);
代码如下:
import java.util.*; public class Main { static final int MAX_N = 10000 + 100; static final int MAX_INT = 100000000; static int[] w = new int[MAX_N]; static int[] c = new int[MAX_N]; static int[] dp = new int[MAX_N]; public static int Min(int a, int b) { return a < b ? a : b; } public static void main(String[] args) { Scanner in = new Scanner(System.in); int case_times = in.nextInt(); while (case_times-- != 0) { int v_pig, v_total, V, N; v_pig = in.nextInt(); v_total = in.nextInt(); N = in.nextInt(); V = v_total - v_pig; Arrays.fill(dp, MAX_INT); dp[0] = 0; for (int i = 1; i <= N; ++ i){ w[i] = in.nextInt(); c[i] = in.nextInt(); } for (int i = 1; i <= N; ++ i) for (int v = c[i]; v <= V; ++ v) dp[v] = Min(dp[v], dp[v - c[i]] + w[i]); if (dp[V] == MAX_INT) System.out.println("This is impossible."); else System.out.println("The minimum amount of money in the piggy-bank is " + dp[V] + "."); } } }
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原文地址:http://www.cnblogs.com/tallisHe/p/4691432.html