B - Piggy-Bank（完全背包）

「爱情、让人受尽委屈。」 2022-03-07 15:10 277阅读 0赞

### 题目描述： ###

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!  
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 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 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 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it’s weight in grams.  
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.”.  
Sample Input  
3  
10 110  
2  
1 1  
30 50  
10 110  
2  
1 1  
50 30  
1 6  
2  
10 3  
20 4  
Sample Output  
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.

##### 题意： #####

在 ACM 能够开展之前，必须准备预算，并获得必要的财力支持。该活动的主要收入来自于 Irreversibly Bound Money (IBM)。思路很简单。任何时候，某位 ACM 会员有少量的钱时，他将所有的硬币投入到小猪储钱罐中。这个过程不可逆，因为只有把小猪储钱罐打碎才能取出硬币。在足够长的时间之后，小猪储钱罐中有了足够的现金，用于支付 ACM 活动所需的花费。  
但是，小猪储钱罐存在一个大的问题，即无法确定其中有多少钱。因此，我们可能在打碎小猪储钱罐之后，发现里面的钱不够。显然，我们希望避免这种不愉快的情况。唯一的可能是，称一下小猪储钱罐的重量，并尝试猜测里面的有多少硬币。假定我们能够精确判断小猪储钱罐的重量，并且我们也知道给定币种的所有硬币的重量。那么，我们可以保证小猪储钱罐中最少有多少钱。  
你的任务是找出最差的情形，即判断小猪储钱罐中的硬币最少有多少钱。我们需要你的帮助。不能再贸然打碎小猪储钱罐了！

###### 分析： ######

比赛的时候，做这道题的时候，一直用的贪心，然后一直过不了。  
～～还是太菜了；  
后来，发现这就是一道完全背包的题目。  
但是又有点区别。这道题是要我们求最少有多少钱。那么我们在递推的时候就需要用min。  
所以dp数组的初始化应该为INF。然后，很明显当i0 j0的时候。dp\[0\]\[0\]=0；  
一定要知道递推公式呀。  
未优化的代码：

#include"stdio.h"
    #include"string.h"
    #include"algorithm"
       using namespace std;
    #define INF 1000000
    typedef struct Node
    {
        int p;
        int w;
    }Node;
    Node node[510];
    int dp[510][10001];
    int main()
    {
        int T;
        int E,F;
        int N;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&E,&F);
            scanf("%d",&N);
            for(int i=0;i<510;i++)
                for(int j=0;j<10001;j++)//初始化为INF
                    dp[i][j]=INF;
            for(int i=1;i<=N;i++)
            {
                scanf("%d%d",&node[i].p,&node[i].w);
            }
            int W=F-E;
            dp[0][0]=0;
            for(int i=1;i<=N;i++)
            {
                for(int j=0;j<=W;j++)
                {
                    int w=node[i].w;
                    int p=node[i].p;
                    if(j>=w)//递推公式
                    {
                        dp[i][j]=min(dp[i-1][j],dp[i][j-w]+p);
                    }
                    else
                        dp[i][j]=dp[i-1][j];
                }
            }
            if(dp[N][W]==INF)
                printf("This is impossible.\n");
            else
                printf("The minimum amount of money in the piggy-bank is %d.\n",dp[N][W]);
        }
    }

优化代码：

#include"stdio.h"
    #include"string.h"
    #include"algorithm"
       using namespace std;
    #define INF 1000000
    typedef struct Node
    {
        int p;
        int w;
    }Node;
    Node node[510];
    int dp[10001];
    int main()
    {
        int T;
        int E,F;
        int N;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&E,&F);
            scanf("%d",&N);
                for(int j=0;j<10001;j++)
                    dp[j]=INF;
            for(int i=1;i<=N;i++)
            {
                scanf("%d%d",&node[i].p,&node[i].w);
            }
            int W=F-E;
            dp[0]=0;
            for(int i=1;i<=N;i++)
            {
                int w=node[i].w;
                int p=node[i].p;
                for(int j=w;j<=W;j++)
                {
                    dp[j]=min(dp[j],dp[j-w]+p);
                }
            }
            if(dp[W]==INF)
                printf("This is impossible.\n");
            else
                printf("The minimum amount of money in the piggy-bank is %d.\n",dp[W]);
        }
    }