P2983 [USACO10FEB]购买巧克力Chocolate Buying

本文介绍了一个关于如何在有限预算内为最多数量的奶牛购买巧克力的问题。通过算法优化选择最经济的巧克力种类来满足最大数量的需求。

题目描述

Bessie and the herd love chocolate so Farmer John is buying them some.

The Bovine Chocolate Store features N (1 <= N <= 100,000) kinds of chocolate in essentially unlimited quantities. Each type i of chocolate has price P_i (1 <= P_i <= 10^18) per piece and there are C_i (1 <= C_i <= 10^18) cows that want that type of chocolate.

Farmer John has a budget of B (1 <= B <= 10^18) that he can spend on chocolates for the cows. What is the maximum number of cows that he can satisfy? All cows only want one type of chocolate, and will be satisfied only by that type.

Consider an example where FJ has 50 to spend on 5 different types of chocolate. A total of eleven cows have various chocolate preferences:

Chocolate_Type Per_Chocolate_Cost Cows_preferring_this_type 1 5 3

2 1 1

3 10 4

4 7 2

5 60 1

Obviously, FJ can't purchase chocolate type 5, since he doesn't have enough money. Even if it cost only 50, it's a counterproductive purchase since only one cow would be satisfied.

Looking at the chocolates start at the less expensive ones, he can * purchase 1 chocolate of type #2 for 1 x 1 leaving 50- 1=49, then * purchase 3 chocolate of type #1 for 3 x 5 leaving 49-15=34, then * purchase 2 chocolate of type #4 for 2 x 7 leaving 34-14=20, then * purchase 2 chocolate of type #3 for 2 x 10 leaving 20-20= 0.

He would thus satisfy 1 + 3 + 2 + 2 = 8 cows.

贝西和其他奶牛们都喜欢巧克力,所以约翰准备买一些送给她们。奶牛巧克力专卖店里

有N种巧克力,每种巧克力的数量都是无限多的。每头奶牛只喜欢一种巧克力,调查显示,

有Ci头奶牛喜欢第i种巧克力,这种巧克力的售价是P。

约翰手上有B元预算,怎样用这些钱让尽量多的奶牛高兴呢?

输入输出格式

输入格式:

 

* Line 1: Two space separated integers: N and B

* Lines 2..N+1: Line i contains two space separated integers defining chocolate type i: P_i and C_i

 

输出格式:

 

* Line 1: A single integer that is the maximum number of cows that Farmer John can satisfy

 

输入输出样例

输入样例#1: 复制
5 50 
5 3 
1 1 
10 4 
7 2 
60 1 
输出样例#1: 复制
8 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c)
{
    return min(min(a, b), c);
}
template <class T> inline T max(T a, T b, T c)
{
    return max(max(a, b), c);
}
template <class T> inline T min(T a, T b, T c, T d)
{
    return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d)
{
    return max(max(a, b), max(c, d));
}
#define scanf1(x) scanf("%d", &x)
#define scanf2(x, y) scanf("%d%d", &x, &y)
#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (int i = a; i <= b; i++)
#define FFor(i, a, b) for (int i = a; i >= b; i--)
#define bug printf("***********\n");
#define mp make_pair
#define pb push_back
const int maxn = 100005;
// name*******************************
ll n,b;
ll ans=0;
struct pp{
ll c,p;
}a[maxn];
bool cmp(pp a,pp b){
return a.p<b.p;
}
// function******************************



//***************************************
int main()
{
//    ios::sync_with_stdio(0);
//    cin.tie(0);
//    freopen("test.txt", "r", stdin);
    //  freopen("outout.txt","w",stdout);
    cin>>n>>b;
    For(i,1,n)
    {
        cin>>a[i].p>>a[i].c;
    }
    sort(a+1,a+1+n,cmp);
    For(i,1,n)
    {
        ll t=b/a[i].p;
        if(t>a[i].c)
        {
            b-=a[i].p*a[i].c;
            ans+=a[i].c;
        }
        else
        {
            ans+=t;
            break;
        }
//        cout<<i<<":"<<ans<<" "<<b<<endl;
    }
    cout<<ans;

    return 0;
}

 

转载于:https://www.cnblogs.com/planche/p/8650449.html

### 关于 USACO 2008 FEB Meteor Shower S 的解题思路 #### 题目概述 题目描述了一种场景,在给定的时间内有若干颗流星会坠落在特定位置并影响周围的区域。贝茜需要找到一条安全路径逃脱,避免被流星击中的同时尽快到达目的地。 #### 解决方法分析 对于此问题的一种有效解决方案是采用广度优先搜索算法(BFS)[^1]。BFS能够帮助探索所有可能的安全移动路线,并记录下最短时间内的逃生路径。具体实现如下: - **初始化队列**:将起点加入到待处理节点列表中。 - **状态表示**:每个状态由当前位置(x,y)及时刻t组成。 - **扩展操作**:从当前状态下尝试向四个方向移动;如果目标格子未受流星影响,则将其作为新的状态压入队列继续考察。 - **终止条件**:一旦发现可以跳出地图边界的位置即视为成功逃离。 为了提高效率,还需要注意以下几点优化措施: - 使用集合来快速判断某个位置是否已经被访问过; - 提前计算好每颗流星的影响范围及其生效时间段,以便高效查询某时刻某处是否安全[^4]。 ```python from collections import deque def bfs(start_x, start_y, meteors): directions = [(0,-1), (-1,0), (0,1), (1,0)] # 初始化受影响时间坐标表 affected_times = [[set() for _ in range(305)] for __ in range(305)] for t,x,y in meteors: for dx,dy in ((0,0),(0,-1),(-1,0),(0,1),(1,0)): nx,ny=x+dx, y+dy if 0<=nx<305 and 0<=ny<305: affected_times[nx][ny].add(t) queue = deque([(start_x,start_y,0)]) visited = set() while queue: cx,cy,time=queue.popleft() if not (0 <= cx < 305 and 0 <= cy < 305): return time if (cx,cy,time) in visited: continue visited.add((cx,cy,time)) next_time=time+1 for dx, dy in directions: nx=cx+dx; ny=cy+dy if next_time not in affected_times[nx][ny]: queue.append((nx,ny,next_time)) return "IMPOSSIBLE" ```
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值