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

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

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.

/*

    看到本题不难想到二维费用的背包f[i][j]=max{f[i-a[i].len][j-a[i].w]+a[i].v}，但是题目要求所有铁轨首尾相连，所以需要对a[]按起点排个序，剩下的就和上面的方程很像了

*/

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

int L,N,B,f[][],ans=-;

struct node{

    int s,t,w,v;

}a[];

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

int main(){

    memset(f,-,sizeof(f));

    f[][]=;

    scanf("%d%d%d",&L,&N,&B);

    for(int i=;i<=N;i++){

        scanf("%d%d%d%d",&a[i].s,&a[i].t,&a[i].v,&a[i].w);

        a[i].t+=a[i].s;

    }

    sort(a+,a+N+,cmp);

    for(int i=;i<=N;i++){

        for(int j=B;j>=a[i].w;j--){

            if(f[a[i].s][j-a[i].w]>=){

                f[a[i].t][j]=max(f[a[i].t][j],f[a[i].s][j-a[i].w]+a[i].v);

            }

        }

    }

    for(int i=;i<=B;i++)ans=max(ans,f[L][i]);

    printf("%d",ans);

}