P2854 [USACO06DEC]牛的过山车Cow Roller Coaster

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.

/*
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);
}

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