• SCOI 2010 股票交易 (单调队列优化dp)


    (f[i][j]) 表示前 (i) 天,当前有 (j) 股的最大值

    啥也不干: (f[i][j] = f[i - 1][j])

    直接买: (j in [0,as[i]] f[i][j] = -j * ap[i])

    (i)(i - w - 1) 的基础上买

    (f[i][j] = max(f[i - w - 1][k] - (j - k) * ap[i]))

    (k in [j - as[i], j])

    (i)(i - w - 1) 的基础上卖

    (f[i][j] = max(f[i - w - 1][k] + (k - j) * bp[i]))

    (k in [j, j + bs[i]])

    显然可以通过单调队列优化,时间复杂度 (O(np))

    #include <map>
    #include <set>
    #include <ctime>
    #include <queue>
    #include <stack>
    #include <cmath>
    #include <vector>
    #include <bitset>
    #include <cstdio>
    #include <cctype>
    #include <string>
    #include <numeric>
    #include <cstring>
    #include <cassert>
    #include <climits>
    #include <cstdlib>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    using namespace std ;
    //#define int long long
    #define rep(i, a, b) for (int i = (a); i <= (b); i++)
    #define per(i, a, b) for (int i = (a); i >= (b); i--)
    #define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
    #define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
    #define clr(a) memset(a, 0, sizeof(a))
    #define ass(a, sum) memset(a, sum, sizeof(a))
    #define lowbit(x) (x & -x)
    #define all(x) x.begin(), x.end()
    #define ub upper_bound
    #define lb lower_bound
    #define pq priority_queue
    #define mp make_pair
    #define pb push_back
    #define pof pop_front
    #define pob pop_back
    #define fi first
    #define se second
    #define iv inline void
    #define enter cout << endl
    #define siz(x) ((int)x.size())
    #define file(x) freopen(#x".in", "r", stdin),freopen(#x".out", "w", stdout)
    typedef long long ll ;
    typedef unsigned long long ull ;
    typedef pair <int, int> pii ;
    typedef vector <int> vi ;
    typedef vector <pii> vii ;
    typedef queue <int> qi ;
    typedef queue <pii> qii ;
    typedef set <int> si ;
    typedef map <int, int> mii ;
    typedef map <string, int> msi ;
    const int N = 2010 ;
    const int INF = 0x3f3f3f3f ;
    const int iinf = 1 << 30 ;
    const ll linf = 2e18 ;
    const int MOD = 1000000007 ;
    const double eps = 1e-7 ;
    void print(int x) { cout << x << endl ; exit(0) ; }
    void PRINT(string x) { cout << x << endl ; exit(0) ; }
    void douout(double x){ printf("%lf
    ", x + 0.0000000001) ; }
    template <class T> void chmin(T &a, T b) { if (a > b) a = b ; }
    template <class T> void chmax(T &a, T b) { if (a < b) a = b ; }
    template <class T> void upd(T &a, T b) { (a += b) %= MOD ; }
    template <class T> void mul(T &a, T b) { a = (ll) a * b % MOD ; }
    
    int n, p, w, ans = -iinf ;
    int ap[N], bp[N], as[N], bs[N], q[N] ;
    int f[N][N] ;
    int hd, tl ;
    
    signed main(){
        scanf("%d%d%d", &n, &p, &w) ;
        rep(i, 1, n) scanf("%d%d%d%d", &ap[i], &bp[i], &as[i], &bs[i]) ;
        ass(f, 0xcf) ;
    	rep(i, 1, n) {
    		rep(j, 0, as[i]) f[i][j] = -j * ap[i] ;
    		rep(j, 0, p) chmax(f[i][j], f[i - 1][j]) ;
    		if (i - w <= 1) continue ;
    		hd = 1, tl = 0 ;
    		rep(j, 0, p) {
    			while (hd <= tl && q[hd] < j - as[i]) hd++ ;
    			if (hd <= tl) { // isn't empty
    				int k = q[hd] ;
    				chmax(f[i][j], f[i - w - 1][k] + (k - j) * ap[i]) ;
    			}
    			while (hd <= tl && f[i - w - 1][q[tl]] + q[tl] * ap[i] <= f[i - w - 1][j] + j * ap[i]) tl-- ;
    			q[++tl] = j ;
    		}
    		hd = 1, tl = 0 ;
    		per(j, p, 0) {
    			while (hd <= tl && q[hd] > j + bs[i]) hd++ ;
    			if (hd <= tl) {
    				int k = q[hd] ;
    				chmax(f[i][j], f[i - w - 1][k] + (k - j) * bp[i]) ;
    			}
    			while (hd <= tl && f[i - w - 1][q[tl]] + q[tl] * bp[i] <= f[i - w - 1][j] + j * bp[i]) tl-- ;
    			q[++tl] = j ;
    		}
    	}
    	rep(i, 0, p) ans = max(ans, f[n][i]) ;
    	printf("%d
    ", ans) ;
    	return 0 ;
    }
    
    /*
    f[i][j] 表示前 i 天,当前有 j 股的最大值
    啥也不干: f[i][j] = f[i - 1][j]
    直接买: j ∈[0, as[i]] f[i][j] = -j * ap[i]
    i 在 i - w - 1 的基础上买:
    f[i][j] = max(f[i - w - 1][k] - (j - k) * ap[i]) (k ∈ [j - as[i], j])
    i 在 i - w - 1 的基础上卖:
    f[i][j] = max(f[i - w - 1][k] + (k - j) * bp[i]) (k ∈ [j, j + bs[i]])
    */
    
    /*
    写代码时请注意:
    	1.ll?数组大小,边界?数据范围?
    	2.精度?
    	3.特判?
    	4.至少做一些
    思考提醒:
    	1.最大值最小->二分?
    	2.可以贪心么?不行dp可以么
    	3.可以优化么
    	4.维护区间用什么数据结构?
    	5.统计方案是用dp?模了么?
    	6.逆向思维?
    */
    
    加油ヾ(◍°∇°◍)ノ゙
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  • 原文地址:https://www.cnblogs.com/harryhqg/p/10621925.html
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