hdu 2126 Buy the souvenirs 01背包+计数dp

Buy the souvenirs

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1647    Accepted Submission(s): 606

Problem 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

  

   2
4 7
1 2 3 4

4 0
1 2 3 4
  

 

Sample Output

  

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

 

Author

wangye

 

Source

HDU 2007-10 Programming Contest

 

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题意：n个物品，m元钱，每个物品最多买一次，问最多可以买几件物品，并且输出方案数。

dp[v]表示体积为1的背包最多能装多少物品，kind[v]表示体积为1的背包装最多物品时，有多少种装发。

G++啊！数组开小了也是wa，wa了好一阵子啊。。

背包中数组的大小视体积而定，活生生被我开成了maxn;

而且，关于计数问题中的初值，要慎重，有时候只有dp[0]=1,其他等于0，

这一题一开始全部要赋值为1。总之要让它能正确的转移啊。

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
#include<cctype>
#include<utility>
#pragma comment(linker, "/STACK:102400000,102400000")
#define PI (4.0*atan(1.0))
#define eps 1e-10
#define sqr(x) ((x)*(x))
#define FOR0(i,n)  for(int i=0 ;i<(n) ;i++)
#define FOR1(i,n)  for(int i=1 ;i<=(n) ;i++)
#define FORD(i,n)  for(int i=(n) ;i>=0 ;i--)
#define  lson   ind<<1,le,mid
#define rson    ind<<1|1,mid+1,ri
#define MID   int mid=(le+ri)>>1
#define zero(x)((x>0? x:-x)<1e-15)
#define mk    make_pair
#define _f     first
#define _s     second
using namespace std;
//const int INF=    ;
typedef long long ll;
//const ll inf =1000000000000000;//1e15;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
const int INF =0x3f3f3f3f;
const int maxn=30+20   ;
const int maxV=510  ;
//const int maxm=    ;

int V,n;
int kind[maxV];
int dp[maxV];
int cost[maxn];
int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&V);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&cost[i]);
        }
        memset(dp,0,sizeof dp);
        //memset(kind,0,sizeof kind);
        //kind[0]=1;
        for(int i=0;i<=V;i++)  kind[i]=1;//我想说这里赋值为1，有个重要的原因是:此时dp[i]都为0，表示价值为0，价值为0的方法当然是一种啊。改变的时候
//也要对应的更改啊！！！！
        for(int i=1;i<=n;i++)
        {
            for(int v=V;v>=cost[i];v--)
            {

                int tmp=dp[v-cost[i]]+1;
                if(tmp>dp[v])        {  kind[v]=kind[v-cost[i]]; dp[v]=tmp;           }
                else if(tmp==dp[v])  {  kind[v]+=kind[v-cost[i]];       }
            }

        }

        if(!dp[V])  puts("Sorry, you can't buy anything.");
        else  printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n",kind[V],dp[V]);





    }


    return 0;
}

/*
2323
5 10
1 2 3 4 5

*/