http://acm.hdu.edu.cn/showproblem.php?pid=3401

Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5188    Accepted Submission(s): 1776


Problem Description
Recently, lxhgww is addicted to stock, he finds some regular patterns after a few days' study.
He forecasts the next T days' stock market. On the i'th day, you can buy one stock with the price APi or sell one stock to get BPi.
There are some other limits, one can buy at most ASi stocks on the i'th day and at most sell BSi stocks.
Two trading days should have a interval of more than W days. That is to say, suppose you traded (any buy or sell stocks is regarded as a trade)on the i'th day, the next trading day must be on the (i+W+1)th day or later.
What's more, one can own no more than MaxP stocks at any time.

Before the first day, lxhgww already has infinitely money but no stocks, of course he wants to earn as much money as possible from the stock market. So the question comes, how much at most can he earn?
 

 

Input
The first line is an integer t, the case number.
The first line of each case are three integers T , MaxP , W .
(0 <= W < T <= 2000, 1 <= MaxP <= 2000) .
The next T lines each has four integers APi，BPi，ASi，BSi( 1<=BPi<=APi<=1000,1<=ASi,BSi<=MaxP), which are mentioned above.
 

 

Output
The most money lxhgww can earn.
 

 

Sample Input
1 5 2 0 2 1 1 1 2 1 1 1 3 2 1 1 4 3 1 1 5 4 1 1
 

 

Sample Output
3
  感觉初始化有点坑，，，，
由于间隔w天才能进行第二次购买，对于第i天想进行交易的股票来说之前的w天如果可以转移说明之前没进行过交易，所以这几天的数值都一样，所以不必重复的更新，
只从前i-w-1这一天转移就好了。
so  f[i][j]=MAX{f[i][j-1], f[i-w-1][k]-(j-k)*ap[i] k<=j&&j-k<=as[i], f[i-w-1][k]+(k-j)*bp[i] | k>=j&&k-j<=bs[i] };
显然可以用单调队列加速，但是注意要提前处理前w天所有购买后的花费， 因为对于i-w-1小于零的天我们直接忽略，导致并没有计算他们，所以切记初始化！
不必考虑前w+1天卖，因为如果想卖说明前面买了，但是两天都处于前w+1天中间隔肯定小于w天，因此证明得前w+1天里不可能发生卖的操作！
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<queue>
 5 #include<algorithm>
 6 using namespace std;
 7 #define inf 120
 8 #define LL long long
 9 inline int read(int f = 1)
10 { 
11     char c = getchar();while (!isdigit(c)) { if (c == '-')f = -1; c = getchar(); }
12     int r = 0; while (isdigit(c)) { r = r * 10 + c - '0';c = getchar(); } return r*f;
13 }
14 int f[2005][2005];
15 int ap[2005], bp[2005], as[2005], bs[2005];
16 struct node2 { int w, k; };
17 deque<node2>q;
18 int main()
19 {
20     int i, j, k, cas, T, MAX_P, W;
21     cin >> cas;
22     while (cas--) {
23         T = read();
24         MAX_P = read();
25         W = read();
26         for (i = 1;i <= T;++i)
27         {
28             ap[i] = read();
29             bp[i] = read();
30             as[i] = read();
31             bs[i] = read();
32         }
33         memset(f, -inf, sizeof(f));
34         for (int i = 1; i <= W + 1; i++) {//第一天到W+1天只都是只能买的
35             for (int j = 0; j <= min(MAX_P, as[i]); j++) {
36                 f[i][j] = -ap[i] * j;
37             }
38         }
39         f[0][0] = 0;
40         for (i = 1;i <= T;++i)
41         {
42             int u = i - W - 1;
43             for (j = 0;j <= MAX_P;++j)
44             {
45                 f[i][j] = max(f[i][j],f[i - 1][j]);
46                 if (u < 0)continue;
47                 while (!q.empty() && q.back().w < f[u][j] + j*ap[i])q.pop_back();
48                 q.push_back(node2{f[u][j]+j*ap[i],j});
49                 while (!q.empty() && j - q.front().k > as[i])q.pop_front();
50                 if (!q.empty()) f[i][j] = max(f[i][j], q.front().w - j*ap[i]);
51             }
52             if (u < 0)continue;
53             q.clear();
54             for (j = MAX_P;j >= 0;--j)
55             {
56                 while (!q.empty() && q.back().w < f[u][j] + j*bp[i])q.pop_back();
57                 q.push_back(node2{f[u][j]+j*bp[i],j});
58                 while (!q.empty() &&  q.front().k -j> bs[i])q.pop_front();
59                 if (!q.empty()) f[i][j] = max(f[i][j], q.front().w - j*bp[i]);
60             }
61             q.clear();
62         }
63         int ans = 0;
64         for (i = 0;i <= MAX_P;++i)
65             ans = max(ans,f[T][i]);
66         printf("%d\n", ans);
67     }
68     return 0;
69 }