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  • 贪心 + DFS

    A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.

    Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.

    You want to buy at least L liters of lemonade. How many roubles do you have to spend?

    Input

    The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

    The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.

    Output

    Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

    Example
    Input
    4 12
    20 30 70 90
    Output
    150
    Input
    4 3
    10000 1000 100 10
    Output
    10
    Input
    4 3
    10 100 1000 10000
    Output
    30
    Input
    5 787787787
    123456789 234567890 345678901 456789012 987654321
    Output
    44981600785557577
    Note

    In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.

    In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.

    In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.

    题意 : 给你 n 个物品,以及一个容器的体积 l , n 个物品的体积是 2^i-1 , 求在超过容器体积的前提下,最小的花费是多少。

    思路分析 : 想了一个贪心策略,优先去贪性价比最高的物品,当恰好装下的时候,此时可以记录一下答案,若不能时,此时可以让他们多装一个,再次记录一下答案,深搜就行了

    代码示例:

    /*
     * Author:  parasol 
     * Created Time:  2018/3/7 18:18:10
     * File Name: 2.cpp
     */
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <string>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <set>
    #include <map>
    #include <time.h>
    using namespace std;
    #define ll long long
    const ll maxn = 1e6+5;
    const double pi = acos(-1.0);
    const ll inf = 0x3f3f3f3f;
    
    struct node
    {
        ll l, c;
        double p;
        bool operator< (const node &v)const{
            return p < v.p;
        }
    }pre[35];
    ll n, l;
    ll ans = __LONG_LONG_MAX__;
    
    void dfs(ll x, ll cost, ll sum){
        if (cost >= ans) return; 
        if (sum <= 0) {ans = min(ans, cost); return;} 
        if (x == n+1) return;
        ll f = sum / pre[x].l;
        int pt = 0;
        if (sum%pre[x].l == 0){
            ans = min(ans, cost+f*pre[x].c);
            return;
        }
        else {
            dfs(x+1, cost+(f+1)*pre[x].c, sum-(f+1)*pre[x].l);
            pt = 1;
        }
        if (pt) {
            dfs(x+1, cost+f*pre[x].c, sum-f*pre[x].l);
        }
    }
    
    int main() {
        //freopen("in.txt", "r", stdin);
        //freopen("out.txt", "w", stdout);
        cin >> n >> l;
        ll an = 1;
        for(ll i = 1; i <= n; i++){
            scanf("%lld", &pre[i].c);
            pre[i].l = an;
            pre[i].p = 1.0*pre[i].c/an;
            an *= 2;
        }
        sort(pre+1, pre+1+n);
        dfs(1, 0, l);
        printf("%lld
    ", ans);
        return 0;
    }
    
    东北日出西边雨 道是无情却有情
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  • 原文地址:https://www.cnblogs.com/ccut-ry/p/8525643.html
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