P2854 [USACO06DEC]牛的过山车Cow Roller Coaster
题目描述
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)．
输入输出格式
输入格式：
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.
输出格式：
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.
输入输出样例
输入样例#1： 复制
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
输出样例#1： 复制
17
说明
Taking the 3rd, 5th and 6th components gives a connected roller-coaster with fun value 17 and cost 7.
Taking the first two components would give a more fun roller-coaster (25) but would be over budget.
思路
- 本题与01背包相似
- 但不同的是添加了对长度与位置的限制
- 那么只要在一维的01背包基础上增加一维表示位置即可
- $f[k,j]$表示使用j钱铺到k的最优值, $g[]$表示欢乐值, $w[]$表示长度, $c[]$表示费用
$$if(x[i]+w[i]=k) f[k,j]=max(f[k-w[i],j-c[i]]+g[i],f[k,j])$$
- 但这样超时，只有x[i]+w[i]=k时才需要考虑转移，那么我们就可以将方程化为
$$f[v][j]=max(f[u][j-c[i]]+g[i],f[v][j]) v=u+w[i]$$
- 但这样就要对数据进行排序才能解决无后效性的问题
代码
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 1324567
using namespace std;
ll l,n,b,ans,f[1002][1002];
struct po{
ll x,w,f,c;
}a[10009];
bool cmp(po xx,po yy){return xx.x<yy.x;}
inline int read(){
int x=0,w=1;
char ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
if(ch=='-') w=-1,ch=getchar();
while(ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-48,ch=getchar();
return x*w;
}
int main()
{
freopen("p2854.in","r",stdin);
freopen("p2854.out","w",stdout);
l=read(),n=read(),b=read();
for(ll 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,cmp);
memset(f,-1,sizeof(f));
ans=-inf;
f[0][0]=0;
for(ll i=1;i<=n;i++) {
ll u=a[i].x;
ll v=a[i].x+a[i].w;
for(ll j=b;j>=a[i].c;j--)
if(f[u][j-a[i].c]!=(-1)) f[v][j]=max(f[u][j-a[i].c]+a[i].f,f[v][j]); //f[u,j]表示用j的钱铺到u的最优值
}
for(ll i=0;i<=b;i++)
ans=max(ans,f[l][i]);
printf("%lld
",ans);
return 0;
}