Codeforces 365D 贪心+dp

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本文探讨了一个物品交换问题，目标是在遵循特定规则的情况下获得最大价值的物品集合。文章详细介绍了问题背景、约束条件及求解思路，并提供了一段C++代码实现。

John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.

John knows that his city has n items in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that each item is one of a kind and that means that you cannot exchange set {a, b} for set {v, a}. However, you can always exchange set x for any set y, unless there is item p, such that p occurs in x and p occurs in y.

For each item, John knows its value ci. John's sense of justice doesn't let him exchange a set of items x for a set of items y, if s(x) + d < s(y) (s(x) is the total price of items in the set x).

During one day John can exchange only one set of items for something else. Initially, he has no items. John wants to get a set of items with the maximum total price. Find the cost of such set and the minimum number of days John can get it in.

Input

The first line contains two space-separated integers n, d (1 ≤ n ≤ 50, 1 ≤ d ≤ 104) — the number of items on the market and John's sense of justice value, correspondingly. The second line contains n space-separated integers ci (1 ≤ ci ≤ 104).

Output

Print two space-separated integers: the maximum possible price in the set of items John can get and the minimum number of days needed to get such set.

Examples
input
3 2
1 3 10
output
4 3
input
3 5
1 2 3
output
6 2
input
10 10000
10000 9999 1 10000 10000 10000 1 2 3 4
output
50010 6

Note

In the first sample John can act like this:

Take the first item (1 - 0 ≤ 2).
Exchange the first item for the second one (3 - 1 ≤ 2).
Take the first item (1 - 0 ≤ 2).

题意：有n个品种的物品，每个物品的价格为c[ i ],每次你可以从你拥有的物品中挑取部分去商店去兑换。兑换有如下限制：

(1)你拿去兑换的物品(不妨设拿去兑换的物品集合为s)，都要换成集合s所没有的。比如(a,b)--->(v,a)就是不合法的

(2)你拿去兑换的物品集s1里的价值和cost1与你兑换回来的物品集s2里的价值和cost2需要满足：cost1+d>=cost2

每天你只能去兑换一次。

问通过兑换可以获得的物品集价值和的最大值以及所需的兑换天数

题目链接：http://codeforces.com/contest/365/problem/D

题解：首先背包枚举下所有可能的状态，然后w表示当前可以到达最大的价值，枚举w+1~w+d可行的状态，转移到最大的可行状态，这里有一点值得思考，为什么转移时会满足条件1？因为从一个低状态w到另一个高状态w+i(i>=1&&i<=d)，因为每种状态我们都只是每个物品最多放1次所到达的，定义状态1: w = a[p_1] + a[p_2] + ... + a[p_k], 状态2: w + i = a[q_1]+a[q_2] + ... + a[q_k'], 2个相同物品，比如a[p_i] = a[q_j], 违反了规则1，我们完全可以保持a[p_i]在w状态里不动，也就是说相同的物品从状态1(低状态)到状态2(高状态)里若存在相同元素，该元素不需调整。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;

#define x0 x0___
#define y0 y0___
#define pb push_back
#define SZ(X) ((int)X.size())
#define mp make_pair
#define Fi first
#define Se second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define ALL(X) X.begin(),X.end()
#define RALL(X) X.rbegin(),X.rend()
#define rep(i,j,k) for(int i = j;i <= k;i ++)
#define per(i,j,k) for(int i = j;i >= k;i --)
#define mem(a,p) memset(a,p,sizeof(a))


const ll MOD = 1E9 + 7;
ll qmod(ll a,ll b,ll c) {ll res=1;a%=c; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%c;a=a*a%c;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

template<typename T, typename S>
void upmax(T& a,S b){if(a<b) a=b;}
template<typename T, typename S>
void upmin(T& a,S b){if(a>b) a=b;}
void gettle() {while(1);}
void getre() {int t=0;t/=t;}


/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
const int N = 57;

int dp[N * 10000];
int a[N];

int main()
{
    int n;
    int d;
    int s = 0;
    scanf("%d %d", &n, &d);
    dp[0] = 1;
    rep (i, 1, n) {
        scanf("%d", &a[i]);
        s += a[i];
        per (j, s, a[i]) {
            dp[j] |= dp[j - a[i]];
        }
    }

    int w, day;
    w = day = 0;

    while(true) {
        bool fg = 0;
        per (i, w + d, w + 1) {
            if(dp[i]) {
                fg = 1;
                day ++;
                w = i;
                break;
            }
        }
        if(!fg) break;
    }

    return !printf("%d %d\n", w, day);
}