2421: Margaritas on the River Walk 给定n件物品和一个背包,第i件物品的体积为Vi,背包容量为C.要求吧一些物品放入背包使得剩下的物品都放不下去,求方案数

2421: Margaritas on the River Walk

One of the more popular activities in San Antonio is to enjoy margaritas in the park along the river know as the River Walk. Margaritas may be purchased at many establishments along the River Walk from fancy hotels to Joe’s Taco and Margarita stand. (The problem is not to find out how Joe got a liquor license. That involves Texas politics and thus is much too difficult for an ACM contest problem.) The prices of the margaritas vary depending on the amount and quality of the ingredients and the ambience of the establishment. You have allocated a certain amount of money to sampling different margaritas.

Given the price of a single margarita (including applicable taxes and gratuities) at each of the various establishments and the amount allocated to sampling the margaritas, find out how many different maximal combinations, choosing at most one margarita from each establishment, you can purchase. A valid combination must have a total price no more than the allocated amount and the unused amount (allocated amount – total price) must be less than the price of any establishment that was not selected. (Otherwise you could add that establishment to the combination.)

For example, suppose you have $25 to spend and the prices (whole dollar amounts) are:

Then possible combinations (with their prices) are:

ABC(25), ABD(24), ABJ(22), ACD(23), ACJ(21), ADJ( 20), AH(24), BCD(24), BCJ(22), BDJ(21), BH(25), CDJ(20), CH(24), DH(23) and HJ(21).

Thus the total number of combinations is 15.

 

Input

The input begins with a line containing an integer value specifying the number of datasets that follow, N (1 <= N <= 1000). Each dataset starts with a line containing two integer values V and D representing the number of vendors (1 <= V <= 30) and the dollar amount to spend (1 <= D <= 1000) respectively. The two values will be separated by one or more spaces. The remainder of each dataset consists of one or more lines, each containing one or more integer values representing the cost of a margarita for each vendor. There will be a total of V cost values specified. The cost of a margarita is always at least one (1). Input values will be chosen so the result will fit in a 32 bit unsigned integer.

Output

For each problem instance, the output will be a single line containing the dataset number, followed by a single space and then the number of combinations for that problem instance.

Sample Input

2
6 25
8 9 8 7 16 5
30 250
1 2 3 4 5 6 7 8 9 10 11
12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30

Sample Output

1 15
2 16509438

 

 

 

 

//

 

 

 

 

借鉴0-1背包
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    int Case=0;
    while(t--)
    {
         int v,d;
         int a[32],sum[32]={0},dp[32][1002]={0};
         scanf("%d%d",&v,&d);
         for(int i=1;i<=v;i++) scanf("%d",&a[i]);
         sort(a+1,a+1+v);
         for(int i=1;i<=v;i++)
            sum[i]=sum[i-1]+a[i];
         for(int i=1;i<=v;i++)
            for(int j=1;j<=d;j++)
                if(j>=sum[i]) dp[i][j]=1;
                else
                {
                     dp[i][j]=dp[i-1][j];
                     if(j-a[i]>=0)
                     {
                         if(dp[i-1][j-a[i]]>0) dp[i][j]+=dp[i-1][j-a[i]];
                         else dp[i][j]++;
                     }
                }
         printf("%d %d/n",++Case,dp[v][d]);
    }
    return 0;
}

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=(1<<25);
int n,v,sum;
int a[11000],f[11000];
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        scanf("%d%d",&n,&v);sum=0;
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        if(v>=sum) {printf("%d 1/n",pl++);continue;}
        if(a[1]>v) {printf("%d 0/n",pl++);continue;}
        sort(a+1,a+1+n);
        int tmp=0;//枚举剩余物品中体积最小的物品：第i个不取，前i-1个取
        for(int i=1;i<=n;tmp+=a[i],i++)
        {
            int m=v-tmp;
            memset(f,0,sizeof(f));f[0]=1;
            //对后面的物品进行DP
            for(int j=i+1;j<=n;j++)
            {
                for(int k=m;k>=a[j];k--)
                {
                    f[k]+=f[k-a[j]];
                }
            }
            for(int j=0;j<=m;j++)
            {
                if(m-j<a[i]) cnt+=f[j];
            }
        }
        printf("%d %d/n", pl++, cnt);
    }
    return 0;
}