zhber 有好多做过的题没写下来，如果我还能记得就补吧

## 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.

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.

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

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
struct work{
int l,r,f,c;
}a;
int n,m,b;
LL f,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;
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);
}

posted on 2014-07-31 21:16  zhber  阅读(132)  评论(0编辑  收藏  举报