HDU 4341 分组背包

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

///
#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=,f=;
    char ch=getchar();
    while(ch<''||ch>'')
    {
        if(ch=='-')f=-;
        ch=getchar();
    }
    while(ch>=''&&ch<='')
    {
        x=x*+ch-'';
        ch=getchar();
    }
    return x*f;
}
//***************************************************************

struct ss
{
    int x,y,v,t;
}p[];
vector<ss >belong[];
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[];
int main()
{

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

代码狗