Description
有 N N 個珠寶 , 每個珠寶價值 CiCi , 能産生 Vi V i 的愉悅度 , 現在你有 M M 元 , 問你最多能獲得多大的愉悅度 , 對于 M∈[1,K]M∈[1,K] 回答問題 .
N≤106,K≤104,Ci≤300 N ≤ 10 6 , K ≤ 10 4 , C i ≤ 300
Solution
首先可以發現 Ci C i 比較小,我們考慮将 Ci C i 的值進行分類,同一類的一起轉移,而且被轉移和轉移到的狀态 modCi mod C i 同餘。對于同一類,我們選擇的珠寶一定是愉悅度從大到小排序後的字首和。
由于是從大到小排序的,是以字首和增長率單調遞減,是以這個DP是單調的,具體證明可以參考肖大佬的部落格:
https://www.cnblogs.com/ShichengXiao/p/9501386.html#autoid-9-0-2
網上似乎都是分治的做法,由于我是沒腦子選手,發現這個可以直接套上二分決策範圍的闆子233,跟詩人小G一樣的做法。
我們對于每一種 Ci C i ,所有同餘的狀态一起處理即可。
Code
/************************************************
* Au: Hany01
* Date: Aug 28th, 2018
* Prob: LOJ6039 雅禮集訓2017Day5 珠寶
* Email: [email protected] & [email protected]
* Inst: Yali High School
************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define ALL(a) a.begin(), a.end()
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia
template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, : ; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, : ; }
inline int read() {
static int _, __; static char c_;
for (_ = , __ = , c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << ) + (_ << ) + (c_ ^ );
return _ * __;
}
const int maxm = + , maxc = ;
int n, m, A[maxm], N, mx, p, top, sz, pos, now, t;
LL dp[maxm], f[][maxm];
PII stk[maxm];
vector<LL> ob[maxc];
inline bool cmp(int x, int y) { return x > y; }
inline LL calc(int i, int j) { return f[now ^ ][A[j]] + (i - j > sz ? : ob[p][i - j - ]); }
int main()
{
#ifdef hany01
freopen("loj6039.in", "r", stdin);
freopen("loj6039.out", "w", stdout);
#endif
static int wi, vi;
n = read(), m = read();
For(i, , n) chkmax(mx, wi = read()), vi = read(), ob[wi].pb(vi);
for (p = ; p <= mx; ++ p) {
if (!(sz = SZ(ob[p]))) continue;
sort(ALL(ob[p]), cmp), now ^= ;
For(i, , sz - ) ob[p][i] += ob[p][i - ];
rep(j, p) {
for (N = , t = j; t <= m; ) A[++ N] = t, dp[N] = f[now ^ ][t], t += p;
stk[top = ] = mp(, );
For(i, , N) {
chkmax(dp[i], calc(i, pos = stk[upper_bound(stk + , stk + + top, mp(i, INF)) - stk - ].y));
if (calc(N, i) >= calc(N, stk[top].y)) {
stk[++ top] = mp(N, i);
for (register int l, r, mid; ; ) {
l = max(stk[top - ].x, i + ), r = N;
while (l < r) {
mid = (l + r) >> ;
if (calc(mid, stk[top - ].y) <= calc(mid, i)) r = mid;
else l = mid + ;
}
stk[top].x = l;
if (stk[top].x > stk[top - ].x) break;
stk[top - ] = stk[top], -- top;
}
}
}
For(i, , N) f[now][A[i]] = dp[i];
}
}
For(i, , m) printf("%lld ", f[now][i]);
return ;
}