「BZOJ1649」[Usaco2006 Dec] Cow Roller Coaster
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.
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.
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
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
选用第3条，第5条和第6条钢轨
题解
f[i][j]表示费用为i长度j的最大乐趣
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#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int l,n,b; struct data{int x,w,f,c;}a[10005]; ll f[1005][1005]; inline bool operator<(data a,data b) { return a.x<b.x; } int main() { memset(f,-1,sizeof(f)); f[0][0]=0; l=read();n=read();b=read(); for(int i=1;i<=n;i++) { a[i].x=read();a[i].w=read();a[i].f=read();a[i].c=read(); } sort(a+1,a+n+1); for(int i=1;i<=n;i++) { int ed=a[i].x+a[i].w; for(int j=a[i].c;j<=b;j++) if(f[j-a[i].c][a[i].x]!=-1)f[j][ed]=max(f[j][ed],f[j-a[i].c][a[i].x]+a[i].f); } ll ans=-1; for(int i=0;i<=b;i++) ans=max(ans,f[i][l]); printf("%lld\n",ans); return 0; } |
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