P2854 [USACO06DEC]牛的过山车Cow Roller Coaster[dp]

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:

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:

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.

题解

这道题还好
具体见代码
d[i][j]表示在0~i都铺满的情况下,成本为j时,有趣值最大为多少
然后就可以当作背包来处理,只是要判断一下。

/*
ID: ylx14271
PROG: telecow
LANG: C++
*/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdlib>
#include<map>
#include<set>
#include<vector>
using namespace std;
int l,n,b;
struct node
{
    int x,w,f,c;
} a[10010];
int d[1010][1010];
//d[i][j]表示在0~i都铺满的情况下,成本为j时,有趣值最大为多少 
bool cmp(node aa,node bb)
{
    if (aa.x==bb.x) return aa.w<bb.w;
    return aa.x<=bb.x;
}
int main()
{
    scanf("%d%d%d",&l,&n,&b);
    for (int i=1;i<=n;i++)
        scanf("%d%d%d%d",&a[i].x,&a[i].w,&a[i].f,&a[i].c),a[i].w+=a[i].x;
    memset(d,-2,sizeof(d));//按照左端大小排序,第二关键字,右端 
    sort(a+1,a+n+1,cmp);//排序是为了后面的操作 
    d[0][0]=0;
    for (int i=1;i<=n;i++)
    {
        for (int j=a[i].c;j<=b;j++)
        if (d[a[i].x][j-a[i].c]>=0) //如果0~a[i].x都铺满了就继续铺 
        d[a[i].w][j]=max(d[a[i].w][j],d[a[i].x][j-a[i].c]+a[i].f);//更新 
    }
    int ans=-1;
    for (int i=1;i<=b;i++) ans=max(d[l][i],ans);
    printf("%d\n",ans);
    return 0;
}