CF----思维排序+01背包山东省第八届省赛K题_历年省选题与cf-优快云博客

本文解析了一个CF风格的比赛问题，涉及动态规划与贪心算法。通过分析比赛规则与得分计算方式，给出了具体的解题思路和排序策略，最终实现01背包算法求解最大得分。

CF

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

LYD loves codeforces since there are many Russian contests. In an contest lasting for T minutes there are n problems, and for the ith problem you can get ai−di∗tipoints, where ai indicates the initial points, di indicates the points decreased per minute (count from the beginning of the contest), and ti stands for the passed minutes when you solved the problem (count from the begining of the contest).
Now you know LYD can solve the ith problem in ci minutes. He can't perform as a multi-core processor, so he can think of only one problem at a moment. Can you help him get as many points as he can?

Input

The first line contains two integers n,T(0≤n≤2000,0≤T≤5000).
The second line contains n integers a1,a2,..,an(0<ai≤6000).
The third line contains n integers d1,d2,..,dn(0<di≤50).
The forth line contains n integers c1,c2,..,cn(0<ci≤400).

Output

Output an integer in a single line, indicating the maximum points LYD can get.

Example Input

3 10
100 200 250
5 6 7
2 4 10

Example Output

Hint

Author

“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）

题目链接：http://www.sdutacm.org/onlinejudge2/index.php/Home/Contest/contestproblem/cid/2081/pid/3903

其实这个题该属于贪心+dp的，比赛的时候死活没想出来。

这个题的意思是说根据cf的规则，我们先定义这场比赛的总时长为T，然后每一个题目有一个分数，有一个增长系数di，解出来这个题目需要ci的时间，现在问你在这场比赛中最多可以获得多少分

首先这个题有顺序关系，所以我们要先确定顺序，我们来排序两个题A和B，如果A在B的前面，损失的分就是（Ca）*da+（Ca+Cb）*db,如果B放在前面就是（Cb）*db+（Cb+Ca）*da，然后根据这个排序就可以了，然后就会发现d是越大越靠前，c是越小越靠前。所以我们用x[i].b/x[i].c排序。

最后我们跑一下01背包取一下最优策略就好了。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
struct node{
    int a,b,c;
    double f;
}x[2009];
int dp[5009];
bool cmp(struct node a,struct node b){
    return a.f>b.f;
}
int main(){
    int n,t;
    scanf("%d%d",&n,&t);
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++)
        scanf("%d",&x[i].a);
    for(int i=1;i<=n;i++)
        scanf("%d",&x[i].b);
    for(int i=1;i<=n;i++)
        scanf("%d",&x[i].c);
    for(int i=1;i<=n;i++)
        x[i].f=1.0*x[i].b/x[i].c;
    sort(x+1,x+n+1,cmp);
    int mix=0;
    for(int i=1;i<=n;i++){
        for(int j=t;j>=x[i].c;j--){
            dp[j]=max(dp[j],dp[j-x[i].c]+x[i].a-x[i].b*j);
            mix=max(dp[j],mix);
        }
    }
    printf("%d\n",mix);
    return 0;
}