HDU1114 Piggy-Bank 【全然背包】

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11149????Accepted Submission(s): 5632

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 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.

題意：給定一個空儲錢罐的重量和滿儲錢罐的重量。再給定一些硬幣種類的價值和重量。每一種硬幣都有無數個。問：在儲錢罐里如何放硬幣能使得總價值最小且總重量正好和滿儲錢罐重量相等。

題解：全然背包入門題。狀態轉移方程為dp[i][j] = min(dp[i-1][j-k*w[i]] + k*v[i]) 0<=k<=totalWeight/w[i],跟01背包一樣，全然背包dp數組也能夠壓縮成一維數組。

僅僅只是內層循環要順序。而01背包是逆序

。

另外這題在初始化dp數組時有個陷阱：對于在函數外定義的數組。若對它使用memset(dp + 1, -1, sizeof(dp + 1));那么將僅僅有一個元素被初始化為-1，其余的全為0.可是若是memset(dp, -1, sizeof(dp));數組就能夠所有初始化為-1。

#include <stdio.h> #include <string.h>int dp[10002];int main() {int totalWeight, weight, val, n, t, i, j;scanf("%d", &t);while(t--){scanf("%d%d", &weight, &totalWeight);totalWeight -= weight;scanf("%d", &n);memset(dp + 1, -1, sizeof(int) * 10001); for(i = 1; i <= n; ++i){scanf("%d%d", &val, &weight);for(j = weight; j <= totalWeight; ++j){if(dp[j - weight] != -1){if(dp[j] == -1 || dp[j] > dp[j - weight] + val)dp[j] = dp[j - weight] + val;}}}if(dp[totalWeight] == -1) printf("This is impossible.\n");else printf("The minimum amount of money in thepiggy-bank is %d.\n",dp[totalWeight]);}return 0; }

posted on 2017-05-14 09:24 mthoutai 閱讀(...) 評論(...) 編輯 收藏

轉載于:https://www.cnblogs.com/mthoutai/p/6851628.html

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