Description
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
Output
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.
題意:確定一個Piggy-bank能不能裝滿,如果可以,求出最少的硬幣值。
分析:完全揹包且揹包要恰好裝滿,涉及到初始化的細節問題,有3種情況:
1: 如果求最大價值且恰好裝滿,則初始化dp[0]爲0,其餘全部爲負無窮大。
2: 如果求最小价值且恰好裝滿,則初始化dp[0]爲0,其餘全部爲正無窮大。
3: 如果不要求全部裝滿,則全部初始化爲0。
方程:dp[j]=max(dp[j],dp[j-w[i]]+v[i]),w[]爲費用,v[]爲價值。
詳見揹包九講......
代碼:
#include<stdio.h>
#include<string.h>
const int inf=10000000;
int dp[10005],v[505],w[505];
int main()
{
int i,j,t,n,a,b;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d%d",&v[i],&w[i]);
dp[0]=0;
int V=b-a;
for(i=1;i<=V;i++)
dp[i]=inf;
for(i=1;i<=n;i++)
for(j=w[i];j<=V;j++)
if(dp[j-w[i]]+v[i]<dp[j])
dp[j]=dp[j-w[i]]+v[i];
if(dp[V]<inf)
printf("The minimum amount of money in the piggy-bank is %d.\n",dp[V]);
else
printf("This is impossible.\n");
}
return 0;
}