(01背包)Buy the souvenirs (hdu 2126)
Buy the souvenirs
Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1904 Accepted Submission(s): 711

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).
In each case, in the first line there are two integer n and m, n is the number of the souvenirs and m is the money you have. The second line contains n integers; each integer describes a kind of souvenir.
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.
4 7
1 2 3 4
4 0
1 2 3 4
Sorry, you can't buy anything.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
using namespace std; const int N = ;
const int INF = 0x3fffffff;
const long long MOD = ;
typedef long long LL;
#define met(a,b) (memset(a,b,sizeof(a))) int a[];
int dp[][N]; /// dp[k][j] 代表选 k 个物品,其中价值为 j 的物品的组合数 int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int i, j, k, n, m, Max=; scanf("%d%d", &n, &m); met(a, );
met(dp, ); for(i=; i<=n; i++)
scanf("%d", &a[i]); dp[][] = ;
for(i=; i<=n; i++)
{
for(k=i; k>=; k--)
{
for(j=a[i]; j<=m; j++)
{ dp[k][j] += dp[k-][j-a[i]];
if(dp[k][j]&&(k>Max)) ///如果 dp[k][j] 有值并且 k>Max 更新Max
Max = k;
}
}
} ///Max 代表从 n 个物品中最多可以选 Max 种物品
///sum 代表有选 Max 个物品的总组合数
int sum = ;
for(i=; i<=m; i++)
sum += dp[Max][i]; if(!Max)
printf("Sorry, you can't buy anything.\n");
else
printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n", sum, Max);
}
return ;
}
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