HDU 4341 分组背包

最新推荐文章于 2025-08-05 13:20:27 发布

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最新推荐文章于 2025-08-05 13:20:27 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zxhl/p/4746136.html

本文解析了一款简化版黄金矿工游戏的算法实现过程，采用分组背包思想解决同一直线上金子的获取顺序问题，确保玩家能在限定时间内获得最大价值。

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B - Gold miner

Time Limit:2000MS

Memory Limit:32768KB

Description

Homelesser likes playing Gold miners in class. He has to pay much attention to the teacher to avoid being noticed. So he always lose the game. After losing many times, he wants your help.

To make it easy, the gold becomes a point (with the area of 0). You are given each gold's position, the time spent to get this gold, and the value of this gold. Maybe some pieces of gold are co-line, you can only get these pieces in order. You can assume it can turn to any direction immediately.
Please help Homelesser get the maximum value.

Input

There are multiple cases.
In each case, the first line contains two integers N (the number of pieces of gold), T (the total time). (0＜N≤200, 0≤T≤40000)
In each of the next N lines, there four integers x, y (the position of the gold), t (the time to get this gold), v (the value of this gold). (0≤|x|≤200, 0＜y≤200,0＜t≤200, 0≤v≤200)

Output

Print the case number and the maximum value for each test case.

Sample Input

3 10 1 1 1 1 2 2 2 2 1 3 15 9 3 10 1 1 13 1 2 2 2 2 1 3 4 7

Sample Output

Case 1: 3 Case 2: 7

题意：

黄金矿工的游戏

同一直线上的金子，只能先抓近的

题解：分组背包

///1085422276
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
typedef long long ll;
using namespace std;
#define inf 10000000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//***************************************************************

struct ss
{
    int x,y,v,t;
}p[2050];
vector<ss >belong[2050];
bool cmp(ss a,ss b)
{
    if(a.y*b.x!=a.x*b.y)
        return a.y*b.x<a.x*b.y;
    else return a.y<b.y;
}
int dp[40005];
int main()
{

    int oo=1;
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)belong[i].clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d%d",&p[i].x,&p[i].y,&p[i].t,&p[i].v);
        }
        p[0].x=-1;
        p[0].y=-1;
        sort(p+1,p+n+1,cmp);
        int zu=0;
        for(int i=1;i<=n;i++)
        {
            if(i!=1&&p[i].y*p[i-1].x==p[i].x*p[i-1].y)
            {
                ss kk;
                kk.v=belong[zu][belong[zu].size()-1].v+p[i].v;
                kk.t=belong[zu][belong[zu].size()-1].t+p[i].t;
                belong[zu].push_back(kk);
            }
            else {
                ss kk;
                kk.v=p[i].v;
                kk.t=p[i].t;
                belong[++zu].push_back(kk);
            }

        }
        //cout<<3213121<<endl;
        for(int i=1;i<=zu;i++)
        {
            for(int j=m;j>=0;j--)
            {
                for(int k=0;k<belong[i].size();k++)
                {
                    if(j>=belong[i][k].t)
                    {
                        dp[j]=max(dp[j],dp[j-belong[i][k].t]+belong[i][k].v);
                    }
                }
            }
        }
        printf("Case %d: %d\n",oo++,dp[m]);

    }

    return 0;
}