BZOJ3163 [Heoi2013]Eden的新背包问题

如果是裸的多重背包就非常简单了。。。

用f[i]表示用了cost为i的时候的最大价值

所以我们二分某一段[l, r]之间的物品不使用，剩下的都使用的最大值，这时的f[]数组可以由之前的f[]数组得到。。。

所以只要记录log(n)的f[]数组即可。。。

 

 1 /**************************************************************
 2     Problem: 3163
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:468 ms
 7     Memory:8512 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13  
14 using namespace std;
15 const int N = 1005;
16 const int Cnt_q = 3e5 + 5;
17 const int Maxlen = N * 15 + Cnt_q * 10;
18  
19 struct query {
20     int del, money, id;
21     query(){}
22     query(int _d, int _m, int _i) : del(_d), money(_m), id(_i) {}
23      
24     inline bool operator < (const query &q) const {
25         return del < q.del;
26     }
27 } q[Cnt_q];
28  
29 int n, mxj, now;
30 int c[N], v[N], t[N];
31 int ans[Cnt_q];
32 int f[15][1005];
33  
34 char buf[Maxlen], *C = buf;
35 int Len;
36  
37 inline int read() {
38     int x = 0;
39     while (*C < '0' || '9' < *C) ++C;
40     while ('0' <= *C && *C <= '9')
41         x = x * 10 + *C - '0', ++C;
42     return x;
43 }
44  
45 void calc(int l, int r, int dep) {
46     int i, j, k, tot;
47     memcpy(f[dep], f[dep - 1], sizeof(f[dep]));
48     for (i = l; i <= r; ++i) {
49         tot = t[i];
50         for (k = 1; k <= tot; k <<= 1) {
51             for (j = mxj; j >= c[i] * k; --j)
52                 f[dep][j] = max(f[dep][j], f[dep][j - c[i] * k] + v[i] * k);
53             tot -= k;
54         }
55         if (!tot) continue;
56         k = tot;
57         for (j = mxj; j >= c[i] * k; --j)
58             f[dep][j] = max(f[dep][j], f[dep][j - c[i] * k] + v[i] * k);
59     }
60 }
61  
62 #define mid (l + r >> 1)
63 void work(int l, int r, int dep) {
64     if (l == r) {
65         for (; q[now].del == l; ++now)
66             ans[q[now].id] = f[dep - 1][q[now].money];
67         return;
68     }
69     calc(mid + 1, r, dep);
70     work(l, mid, dep + 1);
71     calc(l, mid, dep);
72     work(mid + 1, r, dep + 1);
73 }
74 #undef mid
75  
76 int main() {
77     Len = fread(C, 1, Maxlen, stdin);
78     buf[Len] = '\0';
79     int i, x, y, Q;
80     n = read();
81     for (i = 1; i <= n; ++i)
82         c[i] = read(), v[i] = read(), t[i] = read();
83     Q = read();
84     for (i = 1; i <= Q; ++i) {
85         x = read(), y = read();
86         q[i] = query(x + 1, y, i);
87         mxj = max(mxj, y);
88     }
89     sort(q + 1, q + Q + 1), now = 1;
90     work(1, n, 1);
91     for (i = 1; i <= Q; ++i)
92         printf("%d\n", ans[i]);
93     return 0;
94 }
View Code

 

posted on 2015-03-15 19:22  Xs酱~  阅读(711)  评论(0)  编辑  收藏  举报