#
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 504 Solved: 265 [Submit][Status][Discuss]
Description#
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)．
Input#
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.
第1行输入L，N，B，接下来N行，每行四个整数Xi，wi，Fi，Ci．
Output#
- 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.
Sample Input#
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
Sample Output#
17 选用第3条，第5条和第6条钢轨
思路：一个类似01dp的背包
dp[i][j]表示长度为i，成本为j的最大有趣指数。
这个想到了。。
然后转移没有推出来。。。
因为潜意识里总觉得循环的变量是要和dp数组的两维一致。。。
实际上根本无关好吗23333
初始化数组dp为-1，dp[0][0]=0
为-1是因为无解输出-1，这样就不用特别处理了。。。
转移方程为：f[i][end[a]]=max{f[i-cost[a][begin[a]]+w[a]} 当f[i-cost[a][begin[a]]可行时
参考了这篇题解：
1/* ***********************************************
2Author :111qqz
3Created Time :2016年04月11日 星期一 16时43分19秒
4File Name :code/bzoj/1649.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E4+3;
34int n,l,b;
35int dp[1005][1005];
36struct node
37{
38 int l,r;
39 int w;
40 int f;
41 int c;
42
43 bool operator < (node b) const
44 {
45 if (l==b.l) return r<b.r;
46 return l<b.l;
47 }
48
49}p[N];
50int main()
51{
52 #ifndef ONLINE_JUDGE
53 freopen("code/in.txt","r",stdin);
54 #endif
55
56 scanf("%d %d %d",&l,&n,&b);
57 for ( int i = 1 ; i <= n ; i++)
58 {
59 scanf("%d %d %d %d",&p[i].l,&p[i].w,&p[i].f,&p[i].c);
60 p[i].r = p[i].l + p[i].w;
61 }
62
63 sort(p+1,p+n+1);
64
65 ms(dp,-1);
66 dp[0][0] = 0;
67 for ( int i = 1 ; i <= n ; i++)
68 {
69 for ( int j = p[i].c ; j <= b ; j++)
70 if (dp[p[i].l][j-p[i].c]!=-1) dp[p[i].r][j]=max(dp[p[i].r][j],dp[p[i].l][j-p[i].c]+p[i].f);
71 }
72 int ans = -1 ; //无解输出-1
73 for ( int i = 1 ;i <= b ; i++) ans = max(ans,dp[l][i]);
74 printf("%d\n",ans);
75
76
77
78 #ifndef ONLINE_JUDGE
79 fclose(stdin);
80 #endif
81 return 0;
82}