hdu 1114 Piggy-Bank，hdu1114piggy-bank

hdu 1114 Piggy-Bank，hdu1114piggy-bank
Piggy-Bank                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                               Total Submission(s): 11201    Accepted Submission(s): 5657


Problem DescriptionBefore 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! 
 
InputThe 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. 
 
OutputPrint 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.

 
SourceCentral Europe 1999 
簡單的完全背包。。由於求最小值  注意初始化為最大

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define M 100000000using namespace std;struct node{    int w,v;}a[10000];int main(){    int cas,e,f,n,i,j,dp[10000];    cin>>cas;    while(cas--)    {        cin>>e>>f>>n;        int sum=f-e;        for(i=0;i<n;i++)            cin>>a[i].v>>a[i].w;            dp[0]=0;//不能全部初始化為最大            for(i=1;i<=sum;i++)                dp[i]=M;        for(i=0;i<n;i++)            {                for(j=a[i].w;j<=sum;j++)                   {                       dp[j]=min(dp[j],dp[j-a[i].w]+a[i].v);                   }            }            if(dp[sum]==M)                cout<<"This is impossible."<<endl;            else            cout<<"The minimum amount of money in the piggy-bank is "<<dp[sum]<<"."<<endl;    }    return 0;}

杭電ACM題1114把並代碼寫一下給我，最好再寫一下解決的方法說明我懶得寫貌似挺簡單的，幫個忙，謝了

這是個簡單的背包DP哦，建議你先去找那本《電腦演算法設計與分析》的動態規劃部分看看哦，看完0-1背包再看完全背包，這是個完全背包問題。
關於這個轉移方程，書上有類似的解釋，篇幅挺長的，這裡就不寫了
 
杭電acm 1114 怎錯了一道完全背包

確實是完全背包。只是修改了一個地方。
#include<stdio.h>
#include<limits.h>

#define min(a,b) (a)<(b)?(a):(b)
int main()
{
int t,E,F,n;
int p[505],w[505];
int c[505];
unsigned int f[10005]; //這裡，都是 INT_MAX惹得禍。
int v;
int i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&E,&F);
v=F-E;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d%d",&p[i],&w[i]);
for(i=1;i<=v;i++)
f[i]=INT_MAX;
f[0]=0;
for(i=1;i<=n;i++)
for(j=w[i];j<=v;j++)
f[j]=min(f[j],f[j-w[i]]+p[i]);
if(f[v]!=INT_MAX)
printf("The minimum amount of money in the piggy-bank is %d.\n",f[v]);
else
printf("This is impossible.\n");
}
return 0;

}