01背包变形--求方案数

HDU 2126 - Buy the souvenirs

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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.

这道题和01差不多，但是求的是方案数。
“就是0个物品”
”最大值一定是0吧“
“因为只有一种取值”
”0时最大方案数是1 “
“于是后面也就是了”
设dp[j]为j空间下的可选择种类数，a[j]为相应的方案数
可以得到下面的代码：

memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));//这里的初始化很关键for (int i = 0;i <= m;i ++)    a[i] = 1;for (int i = 0;i < n;i ++){    for (int j = m;j >= v[i];j --)    {        int x = dp[j],y = dp[j-v[i]] + 1;        /*x>y也就是说不取第i个的情况（即j空间在前i-1中取）的方案比取第i个，        然后还在前i-1个j-v[i]空间取的种数多，所以不取第i个，跳过*/        if (x > y)  continue;        //x == y  第i个可取可不取        if (x == y)        {            a[j] = a[j] + a[j-v[i]];            continue;        }        /*x<y也就是说要取第i个*/        if (x < y)        {            a[j] = a[j-v[i]];            dp[j] = dp[j-v[i]] + 1;        }    }}

简化之后的代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 30;const int maxm = 500;int t,n,m;int v[maxn+2];int dp[maxm+2],a[maxm+2];int main(){    scanf("%d",&t);    while (t --)    {        scanf("%d%d",&n,&m);        for (int i = 0;i < n;i ++)            scanf("%d",&v[i]);        memset(dp,0,sizeof(dp));        memset(a,0,sizeof(a));        for (int i = 0;i <= m;i ++)            a[i] = 1;        for (int i = 0;i < n;i ++)        {            for (int j = m;j >= v[i];j --)            {                if (dp[j] == dp[j-v[i]] + 1)                {                    a[j] = a[j] + a[j-v[i]];                }                else if (dp[j] < dp[j-v[i]] + 1)                {                    a[j] = a[j-v[i]];                    dp[j] = dp[j-v[i]] + 1;                }            }        }        if (dp[m])              printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n",a[m],dp[m]);        else            printf("Sorry, you can't buy anything.\n");    }    return 0;}