[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[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
const int maxn = 1e5 + 7;
const int maxm = 1e5 + 7;
const int maxv = 1e7 + 7;
const int max_val = 1e6 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-10;

template<class T> T gcd(T a, T b) { return b == 0? a : gcd(b, a % b); }

int f[700], a[40], b[40];

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 = 330;
        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 * 2) {
            if (f[i] >= l) {
                ans = i;
                break;
            }
        }
        cout << ans << endl;
    }
    return 0;
}