HDU 2126 - Buy the souvenirs(01背包)

Description

When the winter holiday comes, a lot of people will have a trip. Generally, there are a lot of souvenirs to sell, and sometimes the travelers will buy some ones with pleasure. Not only can they give the souvenirs to their friends and families as gifts, but also can the souvenirs leave them good recollections. All in all, the prices of souvenirs are not very dear, and the souvenirs are also very lovable and interesting. But the money the people have is under the control. They can’t buy a lot, but only a few. So after they admire all the souvenirs, they decide to buy some ones, and they have many combinations to select, but there are no two ones with the same kind in any combination. Now there is a blank written by the names and prices of the souvenirs, as a top coder all around the world, you should calculate how many selections you have, and any selection owns the most kinds of different souvenirs. For instance:



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).

 

Input

For the first line, there is a T means the number cases, then T cases follow.
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.

 

Output

If you can buy some souvenirs, you should print the result with the same formation as “You have S selection(s) to buy with K kind(s) of souvenirs”, where the K means the most kinds of souvenirs you can buy, and S means the numbers of the combinations you can buy with the K kinds of souvenirs combination. But sometimes you can buy nothing, so you must print the result “Sorry, you can't buy anything.”

 

Sample Input

 24 71 2 3 44 01 2 3 4 

 

Sample Output

 You have 2 selection(s) to buy with 3 kind(s) of souvenirs.Sorry, you can't buy anything.        

                                                                     

题意：

求出最多能够选多少商品，每种商品只能选一次，并求出方案数。

思路：

01背包。dp[j][2] ：dp[j][0] 记录花费为j 是最多商品数， dp[j][1] 记录此时的方案数。

CODE：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;int a[35], dp[505][2];int main(){    //freopen("in", "r", stdin);    int T;    scanf("%d", &T);    while(T--){        memset(dp, 0, sizeof(dp));        int n, m;        scanf("%d %d", &n, &m);        for(int i = 0; i < n; ++i){            scanf("%d", &a[i]);        }        for(int i = 0; i <= m; ++i){//记得初始化，一开始 的方案数都是1.            dp[i][1] = 1;        }        for(int i = 0; i < n; ++i){            for(int j = m; j >= a[i]; --j){                if(dp[j][0] == dp[j - a[i]][0] + 1){                    dp[j][1] = dp[j][1] + dp[j - a[i]][1];                }                else if(dp[j][0] < dp[j - a[i]][0] + 1){                    dp[j][0] = dp[j - a[i]][0] + 1;                    dp[j][1] = dp[j - a[i]][1];                }            }        }        if(dp[m][0] != 0){            printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n", dp[m][1], dp[m][0]);        }        else printf("Sorry, you can't buy anything.\n");    }    return 0;}