hdu 1114 Piggy-Bank

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16946    Accepted Submission(s): 8548

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! 

 

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

310 11021 130 5010 11021 150 301 6210 320 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.
看了别人的0-1背包解法才做出来的，自己按照自己思维的贪心+排序，测试样例都过了，，但是不知道怎么还是错了，有时间看看
错在哪里，，下面先是正确的代码
 #include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define inf 0x3f3f3f3f        //注意dp数组是保存在当前负重w的情况下所能达到的最低钱数，，虽然单个钞票面值<=10000
                                                       但是在有多个的情况下自然不能把10000作为inf，所以inf 需要初始化为一个很大的数using  namespace std;struct Node{    int value;    int weight;}node[501];int  dp[10001];int main(){    int cas,tw;    cin>>cas;    while(cas--)    {        int a,b,kind;        cin>>a>>b;        tw=b-a;        cin>>kind;        for(int i=1;i<=kind;i++)            cin>>node[i].value>>node[i].weight;        for(int i=1;i<=tw;i++)            dp[i]=inf;        dp[0]=0;        for(int i=1;i<=kind;i++)        {            for(int j=node[i].weight;j<=tw;j++)            dp[j]=min(dp[j],dp[j-node[i].weight]+node[i].value);        }        if(dp[tw]!=inf)         printf("The minimum amount of money in the piggy-bank is %d.\n",dp[tw]);        else         printf("This is impossible.\n");    }    return 0;
}
下面 是自己的错误的代码
#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using  namespace std;struct Node{    int value;    int weight;}node[501];int cmp(Node a,Node b){    double xa=a.value/a.weight;    double xb=b.value/b.weight;    return xa<=xb;}int main(){    int cas,tweight;    cin>>cas;    while(cas--)    {        int a,b,kind;        cin>>a>>b;        tweight=b-a;        cin>>kind;        for(int i=1;i<=kind;i++)            cin>>node[i].value>>node[i].weight;        sort(node+1,node+kind+1,cmp);        int temp=tweight,minvalue=0,ge;        for(int i=1;i<=kind;i++)        {            if(node[i].weight<=temp)            {                ge=temp/node[i].weight;                temp%=node[i].weight;                minvalue+=ge*node[i].value;            }        }        if(temp==0)         printf("The minimum amount of money in the piggy-bank is %d.\n",minvalue);        else         printf("This is impossible.\n");    }    return 0;}