HDU1114 Piggy-Bank 完全背包

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26748    Accepted Submission(s): 13524

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.

 

Source

Central Europe 1999

题意：有个存钱罐，通过称重量来估计里面的钱。给定存满时的重量和空的重量，以及各种硬币的重量和面值，求存钱罐里最少的钱数。要求各种硬币重量之和要恰好等于存钱罐中硬币的重量。

解析：这是个完全背包问题，因为每种硬币的数目不限制。不过这个题目要求质量要恰好相等，可以初始化dp全为-1，dp[0] = 0，如果dp[k]不等于-1，才能转移到比k大的数。转移方程：dp[j] = min(dp[j], dp[j-c[i].w]+c[i].p)。

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define N 10005struct Coin{int p, w;};Coin C[505];int dp[N];int main(){int t, e, f, n;scanf("%d", &t);while(t--){scanf("%d%d%d", &e, &f, &n);for(int i = 0; i < n; i++){scanf("%d%d", &C[i].p, &C[i].w);}memset(dp, -1, sizeof(dp));dp[0] = 0;int v = f - e;for(int i = 0; i < n; i++){for(int j = C[i].w; j <= v; j++){int pre = j - C[i].w;if(dp[pre] == -1)continue;if(dp[j] == -1)dp[j] = dp[pre] + C[i].p;elsedp[j] = min(dp[j], dp[pre] + C[i].p);}}if(dp[v] == -1)printf("This is impossible.\n");elseprintf("The minimum amount of money in the piggy-bank is %d.\n", dp[v]);}}