POJ 1384 Piggy-Bank(完全背包问题)

题目链接：http://poj.org/problem?id=1384

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.

Hint:

这是完全背包问题的一个应用，但有所不同的是要求最小值。所以刚开始时dp数组要初始化为一个比较大的整数，如100000000，而自己刚开始时初始化为10000，结果提交了好几次都是WA。所以还要加强练习。另外还要特别注意使用一维数组时，dp[0] 也一定要初始化为0，否则就得不出正确的结果。

代码如下：

#include<iostream>#include<fstream>#include<vector>#include<string>#include<map>#include<iterator>#include<algorithm>#include<numeric>#include<cmath>#include<sstream>#include<bitset>using namespace std;const int MaxN = 501;const int MaxW = 10001;int dp[MaxW + 1];const int MaxInt = 100000000;vector<int> w;vector<int> v;int main(){    #ifdef ONLINE_JUDGE    #else        freopen("D:\\in.txt", "r", stdin);        freopen("D:\\out.txt", "w", stdout);    #endif // ONLINE_JUDEG    int num(0);//总共的case    cin >> num;    int weight_empty(0), weight_full(0);//储蓄罐为空和满时的重量    int n(0);//种类    int weight(0), value(0);    for (int i = 0; i < num; i++)    {        w.clear();        v.clear();        for (int i = 0; i < MaxW + 1; i++)//注意初始化！        {            dp[i] = MaxInt;        }        dp[0] = 0;//注意初始化!用一维数组完全背包来求最小值时，一定要注意把第一个dp值初始化为0        cin >> weight_empty >> weight_full;        int m = weight_full - weight_empty;        cin >>n;        for (int i = 0; i < n; i++)        {            cin >> value >> weight;            v.push_back(value);            w.push_back(weight);        }                for (int i = 0; i <n; i++)        {            for (int j = w[i]; j <= m; j++)            {                dp[j] = min(dp[j], dp[j - w[i]] + v[i]);            }        }        if (dp[m] != MaxInt)        {            cout << "The minimum amount of money in the piggy-bank is "<<dp[m]<<"."<< endl;        }        else        {            cout << "This is impossible." << endl;        }    }    return 0;}