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
输入样例#1：
5 6 10 0 2 20 6 2 3 5 6 0 1 2 1 1 1 1 3 1 2 5 4 3 2 10 2
输出样例#1：
17
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.
代码：
#include<cstdio>
#include<algorithm>
using namespace std;
int idline[1100][11000],f[1100];
int l,n,b;
int ans=-1;
struct node
{
int x,w,f,c; //x起点 w长度 f有趣程度 c花费
}a[11000];
int cmp(node a,node b)
{
return a.f>b.f; //有趣程度从大到小降序
}
int dfs(int id,int money,int funny)
{
if(money>b)
return 0; //花费超出
if(funny<f[a[id].x+a[id].w])//且预算不会超出，且有趣程度比“前辈”（过去到了这个距离的前辈）大
return 0;
f[a[id].x+a[id].w]=funny;
if(a[id].x+a[id].w==l) //到了终点就保存
{
ans=max(ans,funny);
return 0;
}
for(int i=1;i<=idline[a[id].w+a[id].x][0];i++)//以此为起点，开始dfs
{
int start=a[id].w+a[id].x;
dfs(idline[start][i],money+a[idline[start][i]].c,funny+a[idline[start][i]].f);
}
}
int main()
{
scanf("%d%d%d",&l,&n,&b);
for(int i=1;i<=n;i++)
scanf("%d%d%d%d",&a[i].x,&a[i].w,&a[i].f,&a[i].c);
sort(a+1,a+n+1,cmp); //从大到小排序有趣程度
for(int i=1;i<=n;i++)
{
idline[a[i].x][0]++;//计数
idline[a[i].x][idline[a[i].x][0]]=i; //计数每个轨道的起点
}
dfs(0,0,0);
printf("%d\n",ans);
return 0;
}