ACM刷题之路（二十三） HDU 1114 完全背包 Piggy-Bank

亦凉 2024-02-19 20:41 65阅读 0赞

[题目链接：传送门][Link 1]

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 <td> Problem Description Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid. But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs! &nbsp; &nbsp; Input The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 &lt;= E &lt;= F &lt;= 10000. On the second line of each test case, there is an integer number N (1 &lt;= N &lt;= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 &lt;= P &lt;= 50000, 1 &lt;= W &lt;=10000). P is the value of the coin in monetary units, W is it's weight in grams. &nbsp; &nbsp; Output Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.". &nbsp; &nbsp; Sample Input <pre><code>&nbsp;</code></pre> 3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4 &nbsp; &nbsp; Sample Output <pre><code>&nbsp;</code></pre> The minimum amount of money in the piggy-bank is 60. The minimum amount of money in the piggy-bank is 100. This is impossible. </td> 
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## **题意:** ##

**告诉你背包的容量N，和 M 件物品分别各自的体积和质量， 求背包剩余容量为0（即刚好填满）的时候，最小的质量。**

3 三个测试样例

10 110 //N = 110 - 10 = 100 题意为金元宝总质量110 金元宝自己的质量为10 所以物品的质量就是100

2 // 两件物品

1 1 //质量为1 体积为1

30 50 // 质量为30 体积为50

所以**放两个体积50的就满了 质量最少 为 60**

所以输出 The minimum amount of money in the piggy-bank is 60.

1 6

10 3

20 4

第三个例子同理 背包体积为5 物品体积为3 4 ，肯定放不满，所以输出不可能。

## 分析： ##

#### 这个题目是**完全背包的模板题**，可完全背包非常类似于01背包问题，所不同的是每种物品有无限件。也就是从每种物品的角度考虑，与它相关的策略已并非取或不取两种，而是有取0件、取1件、取2件……等很多种。如果仍然按照解01背包时的思路，令f\[i\]\[v\]表示前i种物品恰放入一个容量为v的背包的最大权值。仍然可以按照每种物品不同的策略写出状态转移方程，像这样： ####

f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]|0<=k*c[i]<=v}

**模板是这样的：**

for i=1..N
        for v=0..V
            f[v]=max{f[v],f[v-cost]+weight}

所以只需要套一下模板，注意初始化必须要大一点 99999会WA

**求背包最大重量，初始化0 ，状态转移方程取max**

**求背包最小重量，初始化+∞，状态转移方程取min**

#include<iostream>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 int minn(int x, int y) {
 	return x < y ? x : y;
 }
 const int INF = 99999999;
 struct ss
 {
 	int zhong;
 	int ti;
 }a[502];
 int dp[10002];
 int main()
 {
 	int t;
 	scanf_s("%d", &t);
 	while (t--) {
 		int n, m;
 		scanf_s("%d%d", &n, &m);
 		n = m - n;
 		scanf_s("%d", &m);
 		for (int i = 0; i < m; i++) {
 			scanf_s("%d%d", &a[i].zhong, &a[i].ti);
 		}
 		fill(dp, dp + 10002, INF);
 		dp[0] = 0;
 		for (int i = 0; i < m; i++) {
 			for (int j = a[i].ti; j <= n; j++) {
 				dp[j] = minn(dp[j], dp[j - a[i].ti] + a[i].zhong);
 			}
 		}
 		if (dp[n] == 99999999)
 			puts("This is impossible.");
 		else
 			printf("The minimum amount of money in the piggy-bank is %d.\n", dp[n]);
 	}
 	return 0;
 }

[Link 1]: http://acm.hdu.edu.cn/showproblem.php?pid=1114