[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。

 #pragma comment(linker, "/STACK:10240000,10240000")

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>

 using namespace std;

 #define mem0(a) memset(a, 0, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define rep(a, b) for (int a = 0; a < (b); a++)
 #define rep1(a, b) for (int a = 1; a <= (b); a++)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #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) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pc(a) putchar(a)
 #define ps(a) printf("%s", a)
 #define pd(a) printf("%d", a)
 #define sd(a) scanf("%d", &a)

 typedef double db;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef multiset<int> msi;
 typedef set<int> si;
 typedef vector<int> vi;
 typedef map<int, int> mii;

 const int dx[] = {, , , -, , , -, -};
 const int dy[] = {, , -, , -, , , -};
 const int maxn = 1e5 + ;
 const int maxm = 1e5 + ;
 const int maxv = 1e7 + ;
 const int max_val = 1e6 + ;
 const int MD = 1e9 +;
 const int INF = 1e9 + ;
 const double PI = acos(-1.0);
 const double eps = 1e-;

 template<class T> T gcd(T a, T b) { return b == ? a : gcd(b, a % b); }

 int f[], a[], b[];

 int main() {
     //freopen("in.txt", "r", stdin);
     int n, l;
     while (cin >> n >> l) {
         rep(i, n) {
             sd(a[i]);
             sd(b[i]);
         }
         int maxt = ;
         mem0(f);
         rep(i, n) {
             rep(j, maxt) {
                 f[j + a[i]] = max(f[j + a[i]], f[j] + j * b[i]);
             }
         }
         int ans;
         rep(i, maxt * ) {
             if (f[i] >= l) {
                 ans = i;
                 break;
             }
         }
         cout << ans << endl;
     }
     return ;
 }