动态规划：完全背包问题-HDU1114-Piggy-Bank

解题心得：
1、这是一个完全背包问题的变形，题目要求是求在规定的重量下求价值最小，所以需要将d[0]=0关键的初始化
2、当不可能出现最小的价值时，d的状态并没有被改变，说明并没有放进去一个硬币。

题目：

题目解释：
输入一个空存钱罐的质量，再规定一个存钱罐的存满的质量，告诉你几个硬币的质量和价值，求最小的价值和。

                                    Piggy-Bank

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 24389    Accepted Submission(s): 12367
[/b]

[align=left]Problem Description[/align]
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!

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.".

 

[align=left]Sample Input[/align]

3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4

 

[align=left]Sample Output[/align]

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.

#include<stdio.h>
#include<cstring>
#include<algorithm>
using namespace std;

const int INF = 0x7ffffff;//可以这样规定最大值
struct t
{
int we;
int va;
} ty[510];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int e,f;
scanf("%d%d",&e,&f);
int W = f - e;
int d[10100];
int n;
scanf("%d",&n);
for(int i=1;i<=W;i++)
d[i] = INF;
d[0] = 0;//很关键的初始化
for(int i=0; i<n; i++)
{
scanf("%d%d",&ty[i].va,&ty[i].we);
for(int j=ty[i].we;j<=W;j++)
{
if(d[j] > d[j-ty[i].we] + ty[i].va)
d[j] = d[j-ty[i].we] + ty[i].va;
}
}
if(d[W] == INF)
printf("This is impossible.\n");
else
printf("The minimum amount of money in the piggy-bank is %d.\n",d[W]);
}
return 0;
}