HDU-2126 Buy the souvenirs
数组01背包。
http://acm.hdu.edu.cn/showproblem.php?pid=2126
http://blog.csdn.net/crazy_ac/article/details/7869411
f[i][j][k]表示前i种物品,买了j个,花了小于等于k的钱的时候的方案数
因为是小于等于k,所以初始化的时候要注意哦。
那么转移的时候第i种物品取或者不取
f[i][j][k]+=f[i-1][j][k];
f[i][j][k]+=f[i-1][j-1][k-v[i]];
Buy the souvenirs
Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 748 Accepted Submission(s): 255
And you have only 7 RMB, this time you can select any combination with 3 kinds of souvenirs at most, so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have 8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD (8), and if you have 10 RMB, there is only one selection with the most kinds of souvenirs to you: ABCD (10).All the numbers and results are in the range of 32-signed integer, and 0<=m<=500, 0<n<=30, t<=500, and the prices are all positive integers. There is a blank line between two cases.
#include<iostream>
#include<cstring>
using namespace std;
int f[50][50][510],v[50];
int n,m,ans1,ans2;
int fun()
{
int i,j,k;
for(i=n;i>0;i--)
for(j=i;j>0;j--)
for(k=m;k>=0;k--)
if(f[i][j][k])
{
ans1=f[i][j][k];
ans2=j;
return 1; } return 0;
}
int main()
{
int i,j,k,t;
cin>>t;
while(t--)
{ memset(v,0,sizeof(v));
cin>>n>>m;
for(i=1;i<=n;i++)
cin>>v[i];
memset(f, 0, sizeof(f));
for(i = 0; i <= n; i++)
for(j = 0; j <= m; j++)
f[i][0][j] = 1;
for(i=1;i<=n;i++)
for(j=1;j<=i;j++)
for(k=m;k>=0;k--)
{
f[i][j][k]+=f[i-1][j][k];
if(k>=v[i])
f[i][j][k]+=f[i-1][j-1][k-v[i]];
}
if(fun())
cout<<"You have "<<ans1<<" selection(s) to buy with "<<ans2<<" kind(s) of souvenirs."<<endl;
else
cout<<"Sorry, you can't buy anything."<<endl;
}
return 0;
}
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