BZOJ1855 [Scoi2010]股票交易【单调队列优化dp】

题目链接

BZOJ1855

题解

设\(f[i][j]\)表示第\(i\)天结束时拥有\(j\)张股票时的最大收益

若\(i \le W\)，显然在这之前不可能有交易

\[f[i][j] = max\{f[i - 1][j],-ap[i] * j\} \quad [j \le as[i]]
\]

否则，就有三种选择：

①购买

\[f[i][j] = max\{f[i - W - 1][k] - ap[i] * (j - k)\} \quad[k \le j][j - k \le as[i]]
\]

②卖出

\[f[i][j] = max\{f[i - W - 1][k] + bp[i] * (k - j)\} \quad[k \ge j][k - j \le bs[i]]
\]

③什么也不做

\[f[i][j] = max\{f[i][j],f[i - 1][j]\}
\]

其中③总共是\(O(n^2)\)的

①和②如果逐个枚举是\(O(n^3)\)的，无法承受

拆开式子可发现可以用单调队列优化成\(O(n^2)\)

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define cls(s) memset(s,-0x3f3f3f3f,sizeof(s))

using namespace std;

const int maxn = 2005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int f[maxn][maxn],T,P,W,ap[maxn],bp[maxn],as[maxn],bs[maxn];

struct node{

	int k,v;

}q[maxn];

int head,tail;

int main(){

	T = read(); P = read(); W = read();

	REP(i,T) ap[i] = read(),bp[i] = read(),as[i] = read(),bs[i] = read();

	cls(f); f[0][0] = 0; int ans = 0;

	for (int i = 1; i <= T; i++){

		for (int j = 0; j <= P; j++) f[i][j] = f[i - 1][j];

		if (i <= W){

			for (int j = 0; j <= as[i]; j++)

				f[i][j] = max(f[i][j],-ap[i] * j);

		}

		else {

			head = 0; tail = -1;

			for (int j = 0; j <= P; j++){

				while (head <= tail && (j - q[head].k) > as[i]) head++;

				while (head <= tail && q[tail].v < f[i - W - 1][j] + ap[i] * j) tail--;

				q[++tail] = (node){j,f[i - W - 1][j] + ap[i] * j};

				f[i][j] = max(f[i][j],q[head].v - ap[i] * j);

			}

			head = 0; tail = -1;

			for (int j = P; j >= 0; j--){

				while (head <= tail && (q[head].k - j) > bs[i]) head++;

				while (head <= tail && q[tail].v < f[i - W - 1][j] + bp[i] * j) tail--;

				q[++tail] = (node){j,f[i - W - 1][j] + bp[i] * j};

				f[i][j] = max(f[i][j],q[head].v - bp[i] * j);

			}

		}

		ans = max(ans,f[i][0]);

	}

	printf("%d\n",ans);

	return 0;

}

巴特西

BZOJ1855 [Scoi2010]股票交易【单调队列优化dp】

题目链接

题解

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巴特西

BZOJ1855 [Scoi2010]股票交易 【单调队列优化dp】

题目链接

题解

最新文章

热门文章

BZOJ1855 [Scoi2010]股票交易【单调队列优化dp】