codeforces 183D

本文介绍了一个有趣的问题:如何通过算法确定购买T恤的组合,使得尽可能多的人能得到合适的尺码,采用动态规划的方法解决了该问题,并优化了算法的时间复杂度。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

D. T-shirt
time limit per test5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are going to work in Codeforces as an intern in a team of n engineers, numbered 1 through n. You want to give each engineer a souvenir: a T-shirt from your country (T-shirts are highly desirable in Codeforces). Unfortunately you don’t know the size of the T-shirt each engineer fits in. There are m different sizes, numbered 1 through m, and each engineer will fit in a T-shirt of exactly one size.

You don’t know the engineers’ exact sizes, so you asked your friend, Gerald. Unfortunately, he wasn’t able to obtain the exact sizes either, but he managed to obtain for each engineer i and for all sizes j, the probability that the size of the T-shirt that fits engineer i is j.

Since you’re planning to give each engineer one T-shirt, you are going to bring with you exactly n T-shirts. For those n T-shirts, you can bring any combination of sizes (you can bring multiple T-shirts with the same size too!). You don’t know the sizes of T-shirts for each engineer when deciding what sizes to bring, so you have to pick this combination based only on the probabilities given by your friend, Gerald.

Your task is to maximize the expected number of engineers that receive a T-shirt of his size.

This is defined more formally as follows. When you finally arrive at the office, you will ask each engineer his T-shirt size. Then, if you still have a T-shirt of that size, you will give him one of them. Otherwise, you don’t give him a T-shirt. You will ask the engineers in order starting from engineer 1, then engineer 2, and so on until engineer n.

Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 3000, 1 ≤ m ≤ 300), denoting the number of engineers and the number of T-shirt sizes, respectively.

Then n lines follow, each line contains m space-separated integers. The j-th integer in the i-th line represents the probability that the i-th engineer fits in a T-shirt of size j. Each probability will be given as an integer between 0 and 1000, inclusive. The actual probability should be calculated as the given number divided by 1000.

It is guaranteed that for any engineer, the sum of the probabilities for all m T-shirts is equal to one.

Output
Print a single real number denoting the maximum possible expected number of engineers that will receive a T-shirt.

For the answer the absolute or relative error of 10 - 9 is acceptable.

题意描述:
DescriptionDescription
小A要给N个人准备礼物——T-shirt!但是你不知道他们的尺码……总共有M种尺码,编号从1到M。虽然你不记得每个人准确的尺码,但是你记得对于每一个人i,每一个尺码j,i的尺码正好是j的概率Pij.
现在你要买正好N件T-shirt,求能收到合适尺码礼物的人数的最大期望值。
你送礼物的方式是:从第1个人到第N个人依次询问,如果还有他们的尺码的T-shirt,就送出,否则就不送。
InputInput
第一行有两个整数N,M
接下来有N行,每行有M个整数,第i行的第j个整数表示Pij,用0到1000的整数表示,真实的概率用给出的数字除以1000得到。
保证每一行的整数和是1000。

首先我们可以发现每种礼物(暂且称T恤为礼物)都是独立的,互不影响的

我们考虑第kk种礼物

f[i][j]表示前ii个人有j个人适合的概率
假设pp为概率数组(已经转化为实数)

写出状态转移方程
f[i][j]=f[i1][j1]p[i][k]+f[i1][j](1p[i][k])
但是再把k这一维加进去是不是MLE了啊??
不急我们往下看

同样对于第kk种礼物
g[i]表示买ii件礼物能带来的期望
g[i]=j=0ijf[n][j]+ij=i+1nf[n][j]

这里如果我们不考虑满分,则可直接分组背包合并答案
时间复杂度O(n2m)O(n2m)

这个做法的瓶颈就在于ff数组的大小和合并

我们看g数组的计算形势
随着增大 ,相当于从右边取出一项移到右边。
我们不妨尝试对相邻两项做差,有惊喜!
g[i+1]g[i]=1ij=1f[n][j]g[i+1]−g[i]=1−∑j=1if[n][j]
我们可以发现这个插值是单调的
也就是gg数组是上凸的!
所以g[i]不选g[i+1]g[i+1]也不选

我们考虑贪心
假设我们计算出了所有的g[1]g[1]我们显然取出gg最大的那组物品最优
那我们取出来之后根据相邻两项的差算出下一项,减去前一项的贡献
再和原来的其它g比,就好了
这里我们可以发现ff数组并不是全有用,我们只要随着选的物品增多同步更新f数组即可!
可以省掉一维和大量时间复杂度!
最终时间复杂度:On2+nmO(n2+nm)
对于后面我们可以优化到O(n2+nlogm)O(n2+nlogm)
没有必要这里就不过多篆述

代码如下:

#include<bits/stdc++.h>
using namespace std;

int n , m;
double p[3010][310] , f[310][3010];
double g[3010] , g_lazy[310] , tmp[3010]; 
double ans;
int num[310];
int read(){
    int sum = 0;char c = getchar();bool flag = true;
    while( c < '0' || c > '9' ) {if(c == '-') flag = false;c = getchar();}
    while( c >= '0' && c <= '9' ) sum = sum * 10 + c - 48 , c = getchar();
    if(flag)  return sum;
     else return -sum;
}  

void pre(){
    n = read();m = read();
    for(int i = 1;i <= n;++i)
        for(int j = 1;j <= m;++j)
            p[i][j] = read() * 1.0 / 1000;
    for(int j = 1;j <= m;++j)
        f[j][0] = 1.0;
    for(int i = 1;i <= m;++i){
        for(int j = 1;j <= n;++j)
            f[i][j] = f[i][j - 1] * ( 1 - p[j][i]);
        g_lazy[i] = 1.0 - f[i][n];
    }
    return;
}

void solve(int x){
    num[x]++;
    if(num[x] >= n){
        g_lazy[x] = 0;
        return;
    }
    for(int i = 0;i <= n;++i)
        tmp[i] = f[x][i];
    f[x][0] = 0.0;
    for(int i = 1;i <= n;++i)
        f[x][i] = f[x][i - 1] * (1.0 - p[i][x]) + tmp[i - 1] * p[i][x];
    g_lazy[x] -= f[x][n];
    return;
}

void work(){
    for(int i = 1;i <= n;++i){
        double Max = 0 ;int id = 0;
        for(int j = 1;j <= m;++j)
            if( Max < g_lazy[j] ){
                Max = g_lazy[j];
                id = j;
            }
        ans += Max;
        if(!id) return;
        solve(id);
    }
    printf("%.8lf",ans);
    return;
}

int main(){
    freopen("gift.in","r",stdin);
    freopen("gift.out","w",stdout); 
    pre();
    work();
    return 0;
}
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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