| Trade on Verweggistan |
Since the days of Peter Stuyvesant and Abel Tasman, Dutch merchants have been traveling all over the world to buy and sell goods. Once there was some trade on Verweggistan, but it ended after a short time. After reading this story you will understand why.
At that time Verweggistan was quite popular, because it was the only place in the world where people knew how to make a `prul'. The end of the trade on Verweggistan meant the end of the trade in pruls (or `prullen', as the Dutch plural said), and very few people nowadays know what a prul actually is.
Pruls were manufactured in workyards. Whenever a prul was finished it was packed in a box, which was then placed on top of the pile of previously produced pruls. On the side of each box the price was written. The price depended on the time it took to manufacture the prul. If all went well, a prul would cost one or two florins, but on a bad day the price could easily rise to 15 florins or more. This had nothing to do with quality; all pruls had the same value.
In those days pruls sold for 10 florins each in Holland. Transportation costs were negligible since the pruls were taken as extra on ships that would sail anyway. When a Dutch merchant went to Verweggistan, he had a clear purpose: buy pruls, sell them in Holland, and maximize his profits. Unfortunately, the Verweggistan way of trading pruls made this more complicated than one would think.
One would expect that merchants would simply buy the cheapest pruls, and the pruls that cost more than 10 florins would remain unsold. Unfortunately, all workyards on Verweggistan sold their pruls in a particular order. The box on top of the pile was sold first, then the second one from the top, and so on. So even if the fifth box from the top was the cheapest one, a merchant would have to buy the other four boxes above to obtain it.
As you can imagine, this made it quite difficult for the merchants to maximize their profits by buying the right set of pruls. Not having computers to help with optimization, they quickly lost interest in trading pruls at all.
In this problem, you are given the description of several workyard piles. You have to calculate the maximum profit a merchant can obtain by buying pruls from the piles according to the restrictions given above. In addition, you have to determine the number of pruls he has to buy to achieve this profit.
Input
The input describes several test cases. The first line of input for each test case contains a single integer w, the number of workyards in the test case ( 1
w50).
This is followed by w lines, each describing a pile of pruls. The first number in each line is the number b of
boxes in the pile (0b20).
Following it are b positive integers, indicating the prices (in florins) of the pruls in the stack, given from top to bottom.
The input is terminated by a description starting with w = 0. This description should not be processed.
Output
For each test case, print the case number (1, 2, ...). Then print two lines, the first containing the maximum profit the merchant can achieve. The second line should specify the number of pruls the merchant has to buy to obtain this profit. If this number is not uniquely determined, print the possible values in increasing order. If there are more than ten possible values, print only the 10 smallest.Display a blank line between test cases.
Sample Input
1 6 12 3 10 7 16 5 2 5 7 3 11 9 10 9 1 2 3 4 10 16 10 4 16 0
Sample Output
Workyards 1 Maximum profit is 8. Number of pruls to buy: 4 Workyards 2 Maximum profit is 40. Number of pruls to buy: 6 7 8 9 10 12 13
解題報告：題意有些不好理解，大致是這樣的。商人買貨物，貨物分成一堆一堆的。買某堆的時候必須從頭開始買，如果想買第三個，那麼第一個和第二個必須買。賣的價格都是10塊，問最大利潤，以及要達到最大利潤貨物的數量。數量可能有很多種，打印前10個最小的。
最大利潤很好求。每組求出最大的∑(a[i]), 0 <= i <= b，加起來就是最大的利潤。而數量呢，每組選擇到最大利潤的數量可能有幾種，枚舉每種數量，和之前累積的數量加起來，取10個最小的就好了。這裏用set實現，比較方便。另外使用了C++ 11 新特性，簡易版的枚舉容器類。代碼如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <iomanip>
using namespace std;
#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i>=(m);i--)
#define bit(n) (1LL<<(n))
typedef long long LL;
typedef unsigned long long ULL;
void work();
int main()
{
#ifdef ACM
freopen("in.txt", "r", stdin);
#endif // ACM
work();
}
/***************************************************/
typedef vector<int> vint;
void work()
{
int n;
bool first = true;
int cas = 1;
while(scanf("%d", &n) == 1 && n)
{
if(first)
first = false;
else
puts("");
int ans = 0;
vector<vint> per;
ff(i, n)
{
int b;
scanf("%d", &b);
int maxP = 0;
int sum = 0;
vint price;
ff(j, b)
{
int tmp;
scanf("%d", &tmp);
price.push_back(tmp);
sum += 10 - tmp;
maxP = max(maxP, sum);
}
sum = 0;
vint pos;
if(maxP == 0) pos.push_back(0);
ff(j, b)
{
sum += 10 - price[j];
if(sum == maxP)
pos.push_back(j+1);
}
ans += maxP;
if(pos.size()) per.push_back(pos);
}
printf("Workyards %d\n", cas++);
printf("Maximum profit is %d.\n", ans);
printf("Number of pruls to buy:");
int now = 1, pre = 0;
set<int> ss[2];
ss[now].insert(0);
ff(i, per.size())
{
swap(now, pre);
ss[now].clear();
vint & vec = per[i];
for(int j : vec) for(int v : ss[pre])
{
ss[now].insert(j + v);
}
set<int>::iterator it = ss[now].end();
it--;
while(ss[now].size() > 10)
{
ss[now].erase(it);
it--;
}
}
for(int v : ss[now])
printf(" %d", v);
puts("");
}
}
