题目描述
The cows are building a roller coaster! They want your help to design as fun a roller coaster as possible, while keeping to the budget.
The roller coaster will be built on a long linear stretch of land of length L (1 ≤ L ≤ 1,000). The roller coaster comprises a collection of some of the N (1 ≤ N ≤ 10,000) different interchangable components. Each component i has a fixed length Wi (1 ≤ Wi ≤ L). Due to varying terrain, each component i can be only built starting at location Xi (0 ≤ Xi ≤ L - Wi). The cows want to string together various roller coaster components starting at 0 and ending at L so that the end of each component (except the last) is the start of the next component.
Each component i has a "fun rating" Fi (1 ≤ Fi ≤ 1,000,000) and a cost Ci (1 ≤ Ci ≤ 1000). The total fun of the roller coster is the sum of the fun from each component used; the total cost is likewise the sum of the costs of each component used. The cows' total budget is B (1 ≤ B ≤ 1000). Help the cows determine the most fun roller coaster that they can build with their budget.
奶牛们正打算造一条过山车轨道．她们希望你帮忙，找出最有趣，但又符合预算 的方案． 过山车的轨道由若干钢轨首尾相连，由x=0处一直延伸到X=L(1≤L≤1000)处．现有N(1≤N≤10000)根钢轨，每根钢轨的起点 Xi(0≤Xi≤L- Wi)，长度wi(l≤Wi≤L)，有趣指数Fi(1≤Fi≤1000000)，成本Ci(l≤Ci≤1000)均己知．请确定一 种最优方案，使得选用的钢轨的有趣指数之和最大，同时成本之和不超过B(1≤B≤1000)．
输入输出格式
输入格式：
Line 1: Three space-separated integers: L, N and B.
Lines 2..N+1: Line i+1 contains four space-separated integers, respectively: Xi, Wi, Fi, and Ci.
输出格式：
Line 1: A single integer that is the maximum fun value that a roller-coaster can have while staying within the budget and meeting all the other constraints. If it is not possible to build a roller-coaster within budget, output -1.
输入输出样例
说明
Taking the 3rd, 5th and 6th components gives a connected roller-coaster with fun value 17 and cost 7. Taking the first two components would give a more fun roller-coaster (25) but would be over budget.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c)
{
return min(min(a, b), c);
}
template <class T> inline T max(T a, T b, T c)
{
return max(max(a, b), c);
}
template <class T> inline T min(T a, T b, T c, T d)
{
return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d)
{
return max(max(a, b), max(c, d));
}
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (int i = a; i <= b; i++)
#define FFor(i, a, b) for (int i = a; i >= b; i--)
#define mp make_pair
#define pb push_back
const int maxn = ;
// name*******************************
int f[][];
int L,n,B;
struct node
{
int x,w,f,c;
} a[];
int ans=-;
// function******************************
bool cmp(node a,node b)
{
return a.x<b.x;
} //***************************************
int main()
{
cin>>L>>n>>B;
For(i,,n)
{
cin>>a[i].x>>a[i].w>>a[i].f>>a[i].c;
}
me(f,-);
sort(a+,a++n,cmp);
f[][]=;
For(i,,n)
{
int u=a[i].x;
int v=a[i].x+a[i].w;
FFor(j,B,a[i].c)
{
if(f[u][j-a[i].c]!=-)
f[v][j]=max(f[v][j],f[u][j-a[i].c]+a[i].f);
}
}
For(i,,B)
ans=max(ans,f[L][i]);
cout<<ans; return ;
}