HDU-1114(背包DP)

时间：2015-08-11 00:03:22 阅读：179 评论：0 收藏：0 [点我收藏+]

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Piggy-Bank

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

10 110

1 1

30 50

10 110

1 1

50 30

1 6

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.

分析：题意类似于背包问题，每个钱币都有价值和重量，钱罐总重量告诉你，问你刚好使钱罐装满的最小钱币价值之和是多少？

一开始想过用贪心，即根据每个钱币的价值与重量之比从小到大来装，后来发现wa，于是考虑用dp。写完测试发现怎么也不对，最后原来是没把dp[0]设为零。。。。

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <sstream>
10 #include <queue>
11 #include <typeinfo>
12 #include <fstream>
13 #include <map>
14 #include <stack>
15 using namespace std;
16 #define INF 1000000
17 typedef long long ll;
18 const int maxn=510;
19 struct money{
20     int p,w;
21 }mm[maxn];
22 int dp[10010];
23 int main()
24 {
25     int t;
26     scanf("%d",&t);
27     while(t--){
28         int e,f;
29         scanf("%d%d",&e,&f);
30         int n;
31         scanf("%d",&n);
32         for(int i=0;i<n;i++){
33             scanf("%d%d",&mm[i].p,&mm[i].w);
34         }
35         int tot=f-e;
36         for(int i=1;i<=tot;i++) dp[i]=INF;
37         dp[0]=0;    
38         for(int i=0;i<n;i++){
39             for(int v=mm[i].w;v<=tot;v++){
40                 dp[v]=min(dp[v],dp[v-mm[i].w]+mm[i].p);
41             }
42         }
43         if(dp[tot]!=INF) printf("The minimum amount of money in the piggy-bank is %d.\n",dp[f-e]);
44         else printf("This is impossible.\n");
45     }
46     return 0;
47 }

HDU-1114(背包DP)

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原文地址：http://www.cnblogs.com/RRirring/p/4719572.html

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