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。。。

/**************************************************************
    Problem: 3163
    User: rausen
    Language: C++
    Result: Accepted
    Time:468 ms
    Memory:8512 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
const int N = 1005;
const int Cnt_q = 3e5 + 5;
const int Maxlen = N * 15 + Cnt_q * 10;
 
struct query {
    int del, money, id;
    query(){}
    query(int _d, int _m, int _i) : del(_d), money(_m), id(_i) {}
     
    inline bool operator < (const query &q) const {
        return del < q.del;
    }
} q[Cnt_q];
 
int n, mxj, now;
int c[N], v[N], t[N];
int ans[Cnt_q];
int f[15][1005];
 
char buf[Maxlen], *C = buf;
int Len;
 
inline int read() {
    int x = 0;
    while (*C < '0' || '9' < *C) ++C;
    while ('0' <= *C && *C <= '9')
        x = x * 10 + *C - '0', ++C;
    return x;
}
 
void calc(int l, int r, int dep) {
    int i, j, k, tot;
    memcpy(f[dep], f[dep - 1], sizeof(f[dep]));
    for (i = l; i <= r; ++i) {
        tot = t[i];
        for (k = 1; k <= tot; k <<= 1) {
            for (j = mxj; j >= c[i] * k; --j)
                f[dep][j] = max(f[dep][j], f[dep][j - c[i] * k] + v[i] * k);
            tot -= k;
        }
        if (!tot) continue;
        k = tot;
        for (j = mxj; j >= c[i] * k; --j)
            f[dep][j] = max(f[dep][j], f[dep][j - c[i] * k] + v[i] * k);
    }
}
 
#define mid (l + r >> 1)
void work(int l, int r, int dep) {
    if (l == r) {
        for (; q[now].del == l; ++now)
            ans[q[now].id] = f[dep - 1][q[now].money];
        return;
    }
    calc(mid + 1, r, dep);
    work(l, mid, dep + 1);
    calc(l, mid, dep);
    work(mid + 1, r, dep + 1);
}
#undef mid
 
int main() {
    Len = fread(C, 1, Maxlen, stdin);
    buf[Len] = '';
    int i, x, y, Q;
    n = read();
    for (i = 1; i <= n; ++i)
        c[i] = read(), v[i] = read(), t[i] = read();
    Q = read();
    for (i = 1; i <= Q; ++i) {
        x = read(), y = read();
        q[i] = query(x + 1, y, i);
        mxj = max(mxj, y);
    }
    sort(q + 1, q + Q + 1), now = 1;
    work(1, n, 1);
    for (i = 1; i <= Q; ++i)
        printf("%d
", ans[i]);
    return 0;
}