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题目描述
As it says, Time is Money, Efficiency is Life. A client has a computing task waiting for uploading to the cloud servers. However, due to the resource limits and many other tasks, a task usually cannot be worked on a single server but multiple servers, one after another. You, as an employee of a cloud server provider who specializes in task allocation, need to select several servers from a limited number of cloud servers to perform calculations. We know that even the same servers will perform differently in different environments.
There are n cloud servers available in the vicinity of the client's region numbered from 1 to n. The i-th cloud server has two parameters: wi and pi, which indicate the computing power and transmission efficiency of that server, respectively. The supervisor requires that each task must be computed by exactly m of these servers. The client's computational task is not parallelizable, so you must sequence the work of the m servers to maximize the total computational efficiency. We define the total computational efficiency as follows:
where a1,a2,…,am (pairwise distinct, 1≤ai≤n) are the servers you selected. For convenience, a0=0,p0=1.
输入描述:
The first line contains two integers n,m (1≤n≤10^5, 1≤m≤min(n,20)), denoting the number of cloud servers and servers that you have to select.
The second line contains n integers w1,w2,…,wn(1≤wi≤10^9), denoting the servers' computing power.
The third line contains n integers q1,q2,…,qn, (8000≤qi≤12000), where pi=qi/10000 denotes the i-th server's transmission efficiency.
输出描述:
Output a float number denoting the maximal total computational efficiency. Your answer is considered correct if the relative or absolute error between yours and the standard solution is not greater than 10^−6.
示例1
输入
5 2
1 2 3 4 5
12000 11000 10000 9000 8000
输出
8.5000000000000000
题意：
给出n个任务，要求从其中挑选出m个任务，其中第i个任务的贡献为wi*前面i-1个任务的p值乘积，求最大贡献。
分析：
可以发现m个任务的顺序不同那么最终贡献不同，所以首先需要确定一个最优的顺序，这个顺序可以通过比较交换相邻两项后的贡献差来确定，根据下图的推导过程可以得到贡献差：
之后就可以根据贡献差来sort了，sort后得到的新序列中任意两位置交换一定不会更优，所以只需要在该序列中找到一个长度为m的子序列使其贡献最大，这个过程可以通过dp来实现，一开始想的状态是dp[i][j]表示前i个数中选j个得到的最大贡献，但是这样设状态的话当前状态不能通过上一个状态转移过来，也就是dp[i][j]不一定是通过dp[i-1][j-1]得到的，因为dp[i][j]的值还需要加上一个连乘，很可能加上这个连乘后之前某个状态就比上个状态更优了，所以这道题目不能这么做。
正确的做法是设dp[i][j]表示第i个到第n个任务中选择j个任务得到的最大贡献，这样显然是可以递推得到dp[i][j]的值，也就是dp[i][j] = dp[i+1][j-1]*p[i]+w[i]，不得不说这种构造方法还是很巧妙的！
具体代码如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#define pii pair<int, double>
#define w first
#define p second
using namespace std;
pii a[100005];
double dp[100005][25];//dp[i][j]表示从第i个任务到第n个任务中选j个的最大贡献
bool cmp(pii x, pii y){
return x.w*(1-y.p)+y.w*(x.p-1) > 0;
}
signed main()
{
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++)
scanf("%d", &a[i].w);
for(int i = 1; i <= n; i++){
scanf("%lf", &a[i].p);
a[i].p /= 1e4;
}
sort(a+1, a+n+1, cmp);
dp[n][1] = a[n].w;
for(int i = n-1; i >= 1; i--){
for(int j = 0; j <= m; j++){
//不选
dp[i][j] = dp[i+1][j];
//选
if(j >= 1 && dp[i][j] < dp[i+1][j-1]*a[i].p+a[i].w)
dp[i][j] = dp[i+1][j-1]*a[i].p+a[i].w;
}
}
double ans = 0.0;
printf("%.10f\n", dp[1][m]);
return 0;
}
