[zoj3623]背包模型

给定n种物品，每种物品需要ti时间生产出来，生产出来以后，每单位时间可以创造wi个价值。如果需要创造至少W个价值，求最少时间。

思路：dp[j]表示用时间j所能创造的最大价值，则有转移方程：dp[j + t[i]] = max(dp[j + t[i], dp[j] + t * w[i]])。另外是否需要按一定顺序排序呢??以下是ac代码。

 1 #pragma comment(linker, "/STACK:10240000,10240000")
 2
 3 #include <iostream>
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstdlib>
 7 #include <cstring>
 8 #include <map>
 9 #include <queue>
10 #include <deque>
11 #include <cmath>
12 #include <vector>
13 #include <ctime>
14 #include <cctype>
15 #include <set>
16
17 using namespace std;
18
19 #define mem0(a) memset(a, 0, sizeof(a))
20 #define lson l, m, rt << 1
21 #define rson m + 1, r, rt << 1 | 1
22 #define define_m int m = (l + r) >> 1
23 #define rep(a, b) for (int a = 0; a < (b); a++)
24 #define rep1(a, b) for (int a = 1; a <= (b); a++)
25 #define all(a) (a).begin(), (a).end()
26 #define lowbit(x) ((x) & (-(x)))
27 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
28 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
29 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
30 #define pc(a) putchar(a)
31 #define ps(a) printf("%s", a)
32 #define pd(a) printf("%d", a)
33 #define sd(a) scanf("%d", &a)
34
35 typedef double db;
36 typedef long long LL;
37 typedef pair<int, int> pii;
38 typedef multiset<int> msi;
39 typedef set<int> si;
40 typedef vector<int> vi;
41 typedef map<int, int> mii;
42
43 const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
44 const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
45 const int maxn = 1e5 + 7;
46 const int maxm = 1e5 + 7;
47 const int maxv = 1e7 + 7;
48 const int max_val = 1e6 + 7;
49 const int MD = 1e9 +7;
50 const int INF = 1e9 + 7;
51 const double PI = acos(-1.0);
52 const double eps = 1e-10;
53
54 template<class T> T gcd(T a, T b) { return b == 0? a : gcd(b, a % b); }
55
56 int f[700], a[40], b[40];
57
58 int main() {
59     //freopen("in.txt", "r", stdin);
60     int n, l;
61     while (cin >> n >> l) {
62         rep(i, n) {
63             sd(a[i]);
64             sd(b[i]);
65         }
66         int maxt = 330;
67         mem0(f);
68         rep(i, n) {
69             rep(j, maxt) {
70                 f[j + a[i]] = max(f[j + a[i]], f[j] + j * b[i]);
71             }
72         }
73         int ans;
74         rep(i, maxt * 2) {
75             if (f[i] >= l) {
76                 ans = i;
77                 break;
78             }
79         }
80         cout << ans << endl;
81     }
82     return 0;
83 }