HDU 4341 - Gold miner

B - Gold miner
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU
4341

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

题目模拟黄金矿工的游戏，给出黄金的坐标，每块黄金有自己的耗时和价值，在规定时间内获得最大的价值。因为同一条线上的黄金必须先去上层的才可以去下层的，所以题目转换成分组背包求解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <ctime>
#include <bitset>
using namespace std;

#define N 405
#define F 40005
#define INF 0x3f3f3f3f
#define PB push_back

struct node{
int x,y,t,v;
node(){}
node(int xx,int yy,int tt,int vv):x(xx),y(yy),t(tt),v(vv){}
};

int n,T;
int f[F];

vector<node> a

;

bool cmp(node x,node y){
return x.y<y.y;
}

int gcd(int a,int b){
if (b==0)
return a;
else
return gcd(b,a%b);
}

int main(){
int kase=0;
while (scanf("%d%d",&n,&T)!=EOF){
memset(f,0,sizeof(f));
for (int i=0;i<405;i++)
for (int j=0;j<205;j++)
a[i][j].clear();
int x,y,t,v;
for (int i=0;i<n;i++){
scanf("%d%d%d%d",&x,&y,&t,&v);
int GCD=gcd(abs(x),y);
int xx=x/GCD;
int yy=y/GCD;
xx+=200;
a[xx][yy].PB(node(x,y,t,v));
}

for (int i=0;i<405;i++){
for (int j=0;j<205;j++){
if (a[i][j].size()>0){
sort(a[i][j].begin(),a[i][j].end(),cmp);
for (int k=1;k<a[i][j].size();k++){
a[i][j][k].t+=a[i][j][k-1].t;
a[i][j][k].v+=a[i][j][k-1].v;
}

for (int v=T;v>=0;v--){
for (int k=0;k<a[i][j].size();k++){
if (v>=a[i][j][k].t)
f[v]=max(f[v],f[v-a[i][j][k].t]+a[i][j][k].v);
}
}
}
}
}

printf("Case %d: %d\n",++kase,f[T]);
}

return 0;
}