hdu 1114 Recommend 完全背包

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19428    Accepted Submission(s): 9849

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.

 

题意：小盆友通过往猪猪存钱罐里放钱的方式攒钱做事。存钱罐除非砸坏，否则无法把钱取出。为了知道是否攒了足够的钱，对存钱罐称重。然后告诉每种钱币的重量和价值，问存钱罐里最少可能有多少钱。

完全背包问题描述：有N种物品和一个容量为V的背包，每种物品有无限件。第i种物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大（或最小）

按照0-1背包的思路写出状态转移方程：
f[i][v]=max{f[i-1][v-k*c[i]]+k*w[i]|0<=k*c[i]<=v}

优化后的状态转移方程：
f[i][v]=max{f[i-1][v],f[i][v-c[i]]+w[i]}

实现方法：

for i=1..N
    for v=0..V
        f[v]=max{f[v],f[v-cost]+weight}

完全背包与01背包不同之处在于，每件物品的数量都是无限的。

在处理的时候：01背包是后面的前面dp， 完全背包是从前面到后面dp

我做的第一道完全背包吧～～  唉，菜鸟啊 看了半天～加油吧，孩子～

代码～

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Problem : 1114 ( Piggy-Bank )     Judge Status : Accepted
RunId : 17143578    Language : G++    Author : 1136242673
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<cctype>#include<stdio.h>#define min(a,b)(a<b?a:b)#define max(a,b)(a>b?a:b)#define INF 0x3f3f3f3ftypedef long long ll;#define N 501int p[N],w[N],dp[10001];int main(){ int T,E,F,n,i,j; scanf("%d",&T); while(T--) { scanf("%d%d",&E,&F); int V=F-E; scanf("%d",&n); for(i=0;i<n;i++) scanf("%d%d",&p[i],&w[i]); for(i=0;i<=V;i++) dp[i]=INF; dp[0]=0; for(i=0;i<n;i++) { for(j=w[i];j<=V;j++) { dp[j]=min(dp[j],dp[j-w[i]]+p[i]); } } if(dp[V]==INF) printf("This is impossible.\n"); else printf("The minimum amount of money in the piggy-bank is %d.\n",dp[V]); }}

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<cctype>#include<stdio.h>#define min(a,b)(a<b?a:b)#define max(a,b)(a>b?a:b)#define INF 0x3f3f3f3ftypedef long long ll;#define N 501int p[N],w[N],dp[10001];int main(){   int T,E,F,n,i,j;   scanf("%d",&T);   while(T--)   {       scanf("%d%d",&E,&F);       int V=F-E;  /// 背包容量为装满时减去为空时       scanf("%d",&n);       for(i=0;i<n;i++)       scanf("%d%d",&p[i],&w[i]);       for(i=0;i<=V;i++)           dp[i]=INF;  ///初始化～为最大值       dp[0]=0;       for(i=0;i<n;i++)       {           for(j=w[i];j<=V;j++)  ///简单的完全被包模版～           {               dp[j]=min(dp[j],dp[j-w[i]]+p[i]);           }       }       if(dp[V]==INF)        printf("This is impossible.\n");       else        printf("The minimum amount of money in the piggy-bank is %d.\n",dp[V]);   }}