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

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.

* 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.

* 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.

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

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 #define max(a,b) (a>b?a:b)
 6 
 7 inline void read(int &x)
 8 {
 9     x=0; register char ch=getchar();
10     for(; ch>‘9‘||ch<‘0‘; ) ch=getchar();
11     for(; ch>=‘0‘&&ch<=‘9‘; ch=getchar()) x=x*10+ch-‘0‘;
12 }
13 const int N(10005);
14 int L,n,B,f[1026][1026],ans;
15 struct Node {
16     int l,r,v,c;
17     bool operator < (const Node&x) const
18     {
19         return l<x.l;
20     }
21 }a[N];
22 
23 int Presist()
24 {
25     read(L),read(n),read(B);
26     for(int i=1; i<=n; ++i)
27     {
28         read(a[i].l),read(a[i].r),
29         read(a[i].v),read(a[i].c);
30         a[i].r+=a[i].l;
31     }
32     std::sort(a+1,a+n+1);    ans=-1;
33     memset(f,-1,sizeof(f)); f[0][0]=0;
34     for(int i=1; i<=n; ++i)
35       for(int j=a[i].c; j<=B; ++j)
36         if(f[a[i].l][j-a[i].c]!=-1)
37         f[a[i].r][j]=max(f[a[i].r][j],f[a[i].l][j-a[i].c]+a[i].v);
38     for(int i=1; i<=B; ++i) ans=max(ans,f[L][i]);
39     printf("%d\n",ans);
40     return 0;
41 }
42 
43 int Aptal=Presist();
44 int main(int argc,char**argv){;}