poj 3093 Margaritas on the River Walk(01背包统计)

Margaritas on the River Walk Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2126   Accepted: 792 Description 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 toJoe’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: Vendor A B C D H J Price 8 9 8 7 16 5 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 Hint Note: Some solution methods for this problem may be exponential in the number of vendors. For these methods, the time limit may be exceeded on problem instances with a large number of vendors such as the second example below. Source Greater New York 2006

题目：http://poj.org/problem?id=3093

题意：给你n个物品，和一个总钱数m，求可行方案的总数，也就是买了几件物品，并且钱数不超过范围，还有没购买的物品中的最小价值要比剩余的钱小。。。

分析：可以枚举一个物品不购买，那么价格小于它的物品都要买，价格大于它的物品用01背包统计即可，状态转移方程为

f[i]=sum{ f[i-v[j] ]}j为价格大于当前不买的物品，v[j]<=i ，这个得先按价格排序

这样的复杂度是O(n^2*m)，显然要超时的，貌似数据弱不会超= =

其实优化很简单，就是倒着枚举物品就行，因为每次参与01背包的物品时相似的，只多了一件，所有只需要做一件物品的背包。。

复杂度变成O(n*m)

几组易错的数据

3
29 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
29 2
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
30 466
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

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=1111;
int f[mm],v[mm];
int i,j,k,n,m,c,ans,t,cs=0;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<=m;++i)f[i]=0;
        f[0]=1,c=m;
        for(i=0;i<n;++i)
            scanf("%d",&v[i]),c-=v[i];
        sort(v,v+n);
        ans=(c>=0);
        if(v[0]<=m)
        for(i=n-1;i>=0;--i)
        {
            if((c=c+v[i])>=0)
                for(j=max(0,c-v[i]+1);j<=c;++j)ans+=f[j];
            for(j=m;j>=v[i];--j)f[j]+=f[j-v[i]];
        }
        printf("%d %d\n",++cs,ans);
    }
    return 0;
}