HDU 3401 Trade

Trade

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3401
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

HDOJ Monthly Contest – 2010.05.01

 

解题：单调队列优化dp

 考虑买入

$dp[i][j] = max(dp[i-1][j],dp[i - w - 1][k] + (k - j)\times ap$

如何选择k，发现要维护$dp[i-w-1][k] - (m - k)\times ap$

 

可以根据$dp[i-w-1][k] - (m-k)\times ap + (m - j)\times ap$ 计算出 $dp[i - w - 1][k] + (k - j)\times ap\,k < j$

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2010;
const int INF = 0x3f3f3f3f;
int dp[maxn][maxn],q[maxn],p[maxn],hd,tl;

int main(){
    int kase,ap,bp,as,bs,n,m,w;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d%d",&n,&m,&w);
        for(int i = 1; i <= w + 1; ++i){
            scanf("%d%d%d%d",&ap,&bp,&as,&bs);
            for(int j = 0; j <= m; ++j){
                dp[i][j] = j <= as?-ap*j:-INF;
                if(i > 1) dp[i][j] = max(dp[i][j],dp[i-1][j]);
            }
        }
        for(int i = w + 2; i <= n; ++i){
            scanf("%d%d%d%d",&ap,&bp,&as,&bs);
            int k = i - w - 1;
            hd = tl = 0;
            for(int j = 0; j <= m; ++j){
                int tmp = dp[k][j] - ap*(m - j);
                while(hd < tl && p[tl-1] < tmp) --tl;
                q[tl] = j;
                p[tl++] = tmp;
                while(hd < tl && j - q[hd] > as) ++hd;
                dp[i][j] = max(dp[i-1][j],p[hd] + ap*(m - j));
            }
            hd = tl = 0;
            for(int j = m; j >= 0; --j){
                int tmp = dp[k][j] + bp*j;
                while(hd < tl && p[tl-1] < tmp) --tl;
                q[tl] = j;
                p[tl++] = tmp;
                while(q[hd] - j > bs) ++hd;
                dp[i][j] = max(dp[i][j],p[hd] - bp*j);
            }
        }
        printf("%d\n",dp[n][0]);
    }
    return 0;
}