BZOJ——1649: [Usaco2006 Dec]Cow Roller Coaster

http://www.lydsy.com/JudgeOnline/problem.php?id=1649

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 743  Solved: 370
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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

HINT

 

Source

Silver

因为需要依次连接，以左端点为关键字排序，类似背包的状态转移

 #include <algorithm>
 #include <cstring>
 #include <cstdio>

 #define max(a,b) (a>b?a:b)

 inline void read(int &x)
 {
     x=; register char ch=getchar();
     for(; ch>''||ch<''; ) ch=getchar();
     for(; ch>=''&&ch<=''; ch=getchar()) x=x*+ch-'';
 }
 const int N();
 int L,n,B,f[][],ans;
 struct Node {
     int l,r,v,c;
     bool operator < (const Node&x) const
     {
         return l<x.l;
     }
 }a[N];

 int Presist()
 {
     read(L),read(n),read(B);
     for(int i=; i<=n; ++i)
     {
         read(a[i].l),read(a[i].r),
         read(a[i].v),read(a[i].c);
         a[i].r+=a[i].l;
     }
     std::sort(a+,a+n+);    ans=-;
     memset(f,-,sizeof(f)); f[][]=;
     for(int i=; i<=n; ++i)
       for(int j=a[i].c; j<=B; ++j)
         if(f[a[i].l][j-a[i].c]!=-)
         f[a[i].r][j]=max(f[a[i].r][j],f[a[i].l][j-a[i].c]+a[i].v);
     for(int i=; i<=B; ++i) ans=max(ans,f[L][i]);
     printf("%d\n",ans);
     return ;
 }

 int Aptal=Presist();
 int main(int argc,char**argv){;}