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
题意是有一段0到L的区间,要求用一些线段覆盖,每一条线段都有价值和代价,求在将区间完全覆盖(不能重叠)和总代价不超过B的条件下能获得的最大价值
首先很容易想到一种背包dp的方法
先把线段排序,不用说
f[i][j][k]表示前i个线段覆盖0到j的区间代价为k的最大价值
这样100e肯定TLE+MLE,不用说
然后不会优化,去orz了黄巨大才会做
把第一维省掉,但是依然TLE
但是我们发现枚举第i段线段,那么它能更新的只有几个状态
学着黄巨大把方程倒一下,令f[i][j]表示费用为i,能覆盖0到j的最大价值
那么枚举费用k,因为不能重叠,只会更新从当前线段左端点开始的方案
这样只要O(NB)就可以了
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
struct work{
int l,r,f,c;
}a[10010];
int n,m,b;
LL f[1010][1010],ans=-1;
inline bool cmp(const work &a,const work &b){return a.l<b.l||a.l==b.l&&a.r<b.r;}
inline int max(int a,int b)
{return a>b?a:b;}
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 main()
{
memset(f,-1,sizeof(f));
f[0][0]=0;
n=read();m=read();b=read();
for (int i=1;i<=m;i++)
{
a[i].l=read();
a[i].r=a[i].l+read();
a[i].f=read();
a[i].c=read();
}
sort(a+1,a+m+1,cmp);
for (int i=1;i<=m;i++)
for (int j=a[i].c;j<=b;j++)
if (f[j-a[i].c][a[i].l]!=-1)
f[j][a[i].r]=max(f[j][a[i].r],f[j-a[i].c][a[i].l]+a[i].f);
for (int i=0;i<=b;i++)ans=max(ans,f[i][n]);
printf("%lld\n",ans);
}
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