HDU 1114 Piggy-Bank (完全背包）

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16255 Accepted Submission(s): 8192


[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.

[align=left]Source[/align]
Central Europe 1999

[align=left]Recommend[/align]
Eddy

完全背包，要求恰好装满，如果不能恰好装满，则输出不可能，否则输出最小值。因为是求最小值，恰好装满，所以初始赋值正INF,f[0]=0;

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 10005
#define N 10005
#define INF 0x3f3f3f3f
int f
;
int w
;
int d
;
int n, V, E, F;
int main()
{
int T;
cin >> T;

while (T--)
{
scanf("%d%d", &E, &F);
V = F - E;
scanf("%d", &n);

for (int i = 1; i <= n; i++)
{
scanf("%d%d", &d[i], &w[i]);
}

memset(f, INF, sizeof(f));
f[0] = 0;

for (int i = 1; i <= n; i++)
for (int v = w[i]; v <= V; v++)
{
if(f[v-w[i]]!=INF && f[v-w[i]]+d[i]<f[v])
f[v] = f[v - w[i]] + d[i];
}

if (f[V] != INF)
{
printf("The minimum amount of money in the piggy-bank is %d.\n", f[V]);
}
else
{
printf("This is impossible.\n");
}
}
return 0;
}