The third ACM Shandong Race Pick Apples [greedy + dynamic regulation]

Big backpack problem, just beginning to want to complicate. Segmented filling backpack, a large portion of the front with the most cost-effective fill, the last part of the motion rules can be.

Title Link: http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2408

Pick Apples Time limit:1000ms Memory limit:165536k in doubt. Point here ^_^ Topic Description

Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She was so surprised because she had never seen so many apples before. Each kind of apple have a size and a price to be sold. Now the little girl wants to gain more profits, but she does isn't know how. So she asks to help, and tell she the most profits she can gain. input

The first line there are an integer t (t <=), indicates the number of test cases.
In each case, there is four lines. In the first three lines, there is both integers S and P in each line, which indicates the size (1 <= s<=) and T He price (1 <= P <= 10000) of this kind of apple.

The fourth line there are an integer V, (1 <= V <= 100,000,000) indicates the volume of the girl ' s bag. Output

For each case, first output of the case number then follow the most profits she can gain. Sample Input

11 12 13 16

Sample Output

Case 1:6

#include "iostream" #include "Cstdio" #include "CString" #include "cstdlib" #include "algorithm" typedef long long LL;

using namespace Std;

LL dp[10240];
    struct node{LL sz, pri; double rat;
    BOOL operator < (const NODE & RHS) const {return rat < Rhs.rat;
}
};
    int main (int argc, char const *argv[]) {LL T, Kase = 0; cin >> t;
        while (t--) {Node v[3];
            for (LL i = 0; i < 3; ++i) {cin >> v[i].sz >> V[i].pri;
        V[i].rat = V[i].sz * 1.0/V[I].PRI;
        } sort (V, v + 3);
        memset (DP, 0, sizeof (DP)); LL all;
        Cin >> All;
                    if (All < 10000) {for (ll i = 1, i <= all, ++i) for (ll j = 0; j < 3; ++j)
            if (i-v[j].sz >= 0) Dp[i] = max (Dp[i], Dp[i-v[j].sz] + v[j].pri);
        printf ("Case%LLD:%lld\n", ++kase, Dp[all]); } else{LL cnt = (all-10000)/v[0].SZ;
            All-= cnt * V[0].SZ;
                        for (ll i = 1, i <= all, ++i) for (ll j = 0; j < 3; ++j) if (i-v[j].sz >= 0)
            Dp[i] = max (Dp[i], Dp[i-v[j].sz] + v[j].pri);
        printf ("Case%LLD:%lld\n", ++kase, Dp[all] + cnt * v[0].pri);
}} return 0;
 }