Piggy-Bank (HDU_1114) 完全背包+二进制优化

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18078    Accepted Submission(s): 9132

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.

题目大意:完全背包问题，背包容量为F-E，物品的价值和体积分别为P,W，每个物品的数量可当做无限大，求能否刚好达到储钱罐的重量，并求出能恰好达到的最小价值。

解题思路：完全背包变形，需要将背包恰好存满，并且将求背包的最大值改为求最小值。完全背包的思路跟01背包的思路差别不大，主要是将物品的无限大利用二进制思想分解为有限个，并且将有原本第二重循环（for(int j=volume[i];j<=V;j++)）由逆序改为正序，这样能够达到一件物品可以取多次的目的。状态转移方程为：dp[j]=min(dp[j],dp[j-volume[i]]+value[i])；

二进制优化：

 for(k=0;vol*(pow(2,k))<V;k++,ki++){
value[ki]=val*(pow(2,k));
volume[ki]=vol*(pow(2,k));
}

代码如下：

#include"iostream"#include"cstdio"#include"math.h"#define MAXV 10000+10//最大容量#define MAXK 500*14//做二进制优化后物品的最多种数  N*k  (2^k>10000)#define INF 11000000using namespace std;int dp[MAXV];int value[MAXK];int volume[MAXK];int min(int x,int y){return x<y?x:y;}int main(){int T;int F,E;scanf("%d",&T);while(T--){scanf("%d%d",&E,&F);int V=F-E; //背包容量int n,val,vol;scanf("%d",&n);int ki=1,k=0;for(int i=1;i<=n;i++){//二进制优化存值 scanf("%d%d",&val,&vol);for(k=0;vol*(pow(2,k))<V;k++,ki++){value[ki]=val*(pow(2,k));volume[ki]=vol*(pow(2,k));//printf("%d ",volume[ki]);}}for(int i=1;i<=V;i++){//恰好装满初始化 dp[i]=INF;}dp[0]=0;for(int i=1;i<=ki-1;i++){for(int j=volume[i];j<=V;j++){ //完全背包，正序，这样同一物品可能取多次 dp[j]=min(dp[j],dp[j-volume[i]]+value[i]);} }if(dp[V]<INF)printf("The minimum amount of money in the piggy-bank is %d.\n",dp[V]);elseprintf("This is impossible.\n");}return 0;}