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!
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
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 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.
这道题要做的就是两件事：
（一）、求出所给的存钱罐中的质量（满存钱罐的值-空存钱罐的值）能否被所给硬币填满，若能，则证明可以被输出值，不然就是impossible！
（二）、在（一）存在条件下，我们去求满存钱罐时的最小值，也就是最坏情况，用一个完全背包即可。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int idx=1e9+5;
int coin_w; //硬币的实际总质量
int N;
int p[505],w[505]; //p是货币值，w是质量
int dp[10005];
int wp[10005];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
for(int i=0; i<10004; i++)
{
dp[i]=idx;
}
dp[0]=0;
memset(wp, 0, sizeof(wp));
int r1,r2;
scanf("%d%d",&r1,&r2);
coin_w=r2-r1;
scanf("%d",&N);
for(int i=0; i<N; i++)
{
scanf("%d%d",&p[i],&w[i]);
}
for(int i=0; i<N; i++)
{
for(int j=w[i]; j<=coin_w; j++)
{
if(dp[j]>dp[j-w[i]]+p[i])
{
dp[j]=dp[j-w[i]]+p[i];
}
}
}
for(int i=0; i<N; i++)
{
for(int j=w[i]; j<=coin_w; j++)
{
if(wp[j]<wp[j-w[i]]+w[i])
{
wp[j]=wp[j-w[i]]+w[i];
}
}
}
if(wp[coin_w]==coin_w)
{
printf("The minimum amount of money in the piggy-bank is %d.\n",dp[coin_w]);
}
else
{
printf("This is impossible.\n");
}
}
return 0;
}
最近有了些特殊的想法，对于代码做了些小改动，这样就不用再花一个数组去记录重量了：
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
int real_weight;
int N;
struct node
{
int v,w;
}a[505];
int dp[10005];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(a, 0, sizeof(a));
memset(dp, 0, sizeof(dp));
int e1,e2;
scanf("%d%d",&e1,&e2);
real_weight=e2-e1;
scanf("%d",&N);
for(int i=0; i<N; i++)
{
scanf("%d%d",&a[i].v,&a[i].w);
}
for(int i=0; i<N; i++)
{
for(int j=a[i].w; j<=real_weight; j++)
{
if(dp[j-a[i].w] || !(j-a[i].w))
{
if(!dp[j])dp[j]=dp[j-a[i].w]+a[i].v;
dp[j]=min(dp[j], dp[j-a[i].w]+a[i].v);
}
}
}
if(dp[real_weight]) printf("The minimum amount of money in the piggy-bank is %d.\n",dp[real_weight]);
else printf("This is impossible.\n");
}
return 0;
}
