HDU 3591【多重背包】

题目：The trouble of Xiaoqian

题意：

有N种coin,给出每种coin的价值Vi,和小强拥有的个数Ci,小强去购物，要付m元，求小强和店员间交换货币时的最小个数货币，即小强付出coin个数加上店员找回coin个数。

解题思路:

背包题，先用二进制思想把coin合并成和种不同价值的货币，再进行01背包。

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <map>
 8 
 9 using namespace std;
10 const int MAXN = 100 + 10;
11 const int INF = 0x3f7f7f7f; 
12 int V[MAXN];
13 int C[MAXN];
14 
15 struct TCoin
16 {
17     int val, num;
18 }Coin[MAXN * MAXN * MAXN];
19 
20 int DP[20000 + 10];
21 int main()
22 {
23     freopen("in.txt","r",stdin);
24     int N, T;
25     int caseNumber = 1;
26     while(scanf("%d%d", &N, &T) == 2 && (N || T))
27     {
28         printf("Case %d: ", caseNumber++);
29         for(int i = 1; i <= N; ++i)
30             scanf("%d", &V[i]);
31         for(int i = 1; i <= N; ++i)
32             scanf("%d", &C[i]);
33         int end = 1;
34         for(int i = 1, n; i <= N; ++i)
35         {
36             for(n = 1; n <= C[i]; n <<= 1)
37             {
38                 Coin[end].val = n * V[i];
39                 Coin[end].num = n;
40                 ++end;
41             }
42             n >>= 1;
43             if(n < C[i])
44             {
45                 Coin[end].val = (C[i] - n) * V[i];
46                 Coin[end].num = C[i] - n;
47                 ++end;
48             }
49         }
50         for(int i = 1; i <= 20000; ++i)
51             DP[i] = INF;
52         DP[0] = 0;
53         for(int i = 1; i < end; ++i)
54         {
55             for(int j = 20000; j >= Coin[i].val; --j)
56             {
57                 if(DP[j] > DP[j - Coin[i].val] + Coin[i].num)
58                     DP[j] = DP[j - Coin[i].val] + Coin[i].num;
59             }
60         }
61         int ans = INF;
62         for(int i = T; i <= 20000; ++i)
63             if(DP[i] + DP[i - T] < ans)
64                 ans = DP[i] + DP[i - T];
65         if(ans == INF)
66             printf("-1\n");
67         else
68             printf("%d\n", ans);
69     }
70     return 0;
71 }