hdu1114 Piggy-Bank 求最小值的背包
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Piggy-Bank
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4603 Accepted Submission(s): 2269
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!
题意:一个储钱罐,知道里面硬币的重量,现在提供N种类型的硬币的重量和价格(数目无限),求储钱罐里面最少的价值。
即:求出把背包装满可以获得的最小的价值.
唯一的区别就是初始化
该题利用了我们的逆向思维,同时要注意该题他的质量是一定的,也就是说背包一定要是满的,刚开始对于这类背包我们令初始值是负无穷,而这题则相反,令初始值是正无穷,每次区最小的数,
同时要注意f[j-weight[i]]!=inf应为一相等就与背包一定要满的条件相矛盾; 但是这个条件我没有去用 会稍微浪费点时间把
因为要正好填满 所以必须要初始化很大
#include<stdio.h>
int val[600],wei[600],bag[10005];
int _bag(int v,int n)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=wei[i];j<=v;j++)
{
bag[j]=(bag[j-wei[i]]+val[i])<bag[j]?(bag[j-wei[i]]+val[i]):bag[j];
// printf("%d ",bag[j]);
}
// printf("\n");
}
return bag[v];
}
int main()
{
int t,i,n,v,e,f,ans;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&e,&f);
v=f-e;
for(i=0;i<v+3;i++)
bag[i]=999999999;
bag[0]=0;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d %d",&val[i],&wei[i]);
ans=_bag(v,n);
if(ans==999999999) printf("This is impossible.\n");
else printf("The minimum amount of money in the piggy-bank is %d.\n",ans);
}
return 0;
}
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