I love sneakers!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4464 Accepted Submission(s): 1824
Problem Description
After months of hard working, Iserlohn finally wins awesome amount of scholarship. As a great zealot of sneakers, he decides to spend all his money on them in a sneaker store.
There are several brands of sneakers that Iserlohn wants to collect, such as Air Jordan and Nike Pro. And each brand has released various products. For the reason that Iserlohn is definitely a sneaker-mania, he desires to buy at least one product for each brand.
Although the fixed price of each product has been labeled, Iserlohn sets values for each of them based on his own tendency. With handsome but limited money, he wants to maximize the total value of the shoes he is going to buy. Obviously, as a collector, he won’t buy the same product twice.
Now, Iserlohn needs you to help him find the best solution of his problem, which means to maximize the total value of the products he can buy.
Input
Input contains multiple test cases. Each test case begins with three integers 1<=N<=100 representing the total number of products, 1 <= M<= 10000 the money Iserlohn gets, and 1<=K<=10 representing the sneaker brands. The following N lines each represents a product with three positive integers 1<=a<=k, b and c, 0<=b,c<100000, meaning the brand’s number it belongs, the labeled price, and the value of this product. Process to End Of File.
Output
For each test case, print an integer which is the maximum total value of the sneakers that Iserlohn purchases. Print "Impossible" if Iserlohn‘s demands can’t be satisfied.
Sample Input
5 10000 3
1 4 6
2 5 7
3 4 99
1 55 77
2 44 66
Sample Output
255
Source
2009 Multi-University Training Contest 13 - Host by HIT
题目意思：
有n种物品，每种可能有多个，每个有不同体积和价值。初始手中有V的背包，现将所有种类至少一个装进背包中，若无法达到目标输出Impossible，否则输出最大的价值。
思路：
原始的分组背包问题是每种选或不选，而且选的话只能选一个。而这道题是每种必选，选的话可以选多个。
选多个的条件很容易，把分组背包for V 和 for 第i个 互换一下层数即可。
必选的条件有点难度，若必选的话，那么dp[i][j]由dp[i][j-v[k]]和dp[i-1][j-v[k]]转移即 在第i组中去掉v[k]体积和在前i-1组中去掉v[k]体积，在取max时不能用max(dp[i-1][j],dp[i-1][j-v[k]]+w[k])，因为这种max的意思是这组可能不选任何一个，所以需要用max(dp[i][j],dp[i-1][j-v[k]+w[k])。
初始dp为-1，当dp=-1时这个状态是不可行的，不能转移，这个是判断是否能达到目标。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 #include <cmath>
8 #include <set>
9 using namespace std;
10
11 #define N 105
12
13 int max(int x,int y){return x>y?x:y;}
14 int min(int x,int y){return x<y?x:y;}
15 int abs(int x,int y){return x<0?-x:x;}
16
17
18 struct node{
19 int v, w;
20 };
21
22 int dp[15][10005];
23 vector<node>ve[15];
24 int n;
25
26 main()
27 {
28 int i, j, k;
29 node p;
30 int V;
31 while(scanf("%d %d %d",&n,&V,&k)==3){
32 for(i=0;i<=k;i++) ve[i].clear();
33 for(i=0;i<n;i++){
34 scanf("%d %d %d",&j,&p.v,&p.w);
35 ve[j].push_back(p);
36 }
37 memset(dp,-1,sizeof(dp));
38 memset(dp[0],0,sizeof(dp[0]));
39 for(i=1;i<=k;i++){
40 for(j=0;j<ve[i].size();j++){
41 p=ve[i][j];
42 for(int v=V;v>=p.v;v--){
43 if(dp[i][v-p.v]!=-1) dp[i][v]=max(dp[i][v],dp[i][v-p.v]+p.w);
44 if(dp[i-1][v-p.v]!=-1) dp[i][v]=max(dp[i][v],dp[i-1][v-p.v]+p.w);
45 }
46 }
47 }
48 if(dp[k][V]==-1) printf("Impossible\n");
49 else printf("%d\n",dp[k][V]);
50 }
51 }