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
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
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.
/* 看到本题不难想到二维费用的背包f[i][j]=max{f[i-a[i].len][j-a[i].w]+a[i].v},但是题目要求所有铁轨首尾相连,所以需要对a[]按起点排个序,剩下的就和上面的方程很像了 */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int L,N,B,f[1010][1010],ans=-1; struct node{ int s,t,w,v; }a[10010]; bool cmp(node x,node y){return x.s<y.s;} int main(){ memset(f,-1,sizeof(f)); f[0][0]=0; scanf("%d%d%d",&L,&N,&B); for(int i=1;i<=N;i++){ scanf("%d%d%d%d",&a[i].s,&a[i].t,&a[i].v,&a[i].w); a[i].t+=a[i].s; } sort(a+1,a+N+1,cmp); for(int i=1;i<=N;i++){ for(int j=B;j>=a[i].w;j--){ if(f[a[i].s][j-a[i].w]>=0){ f[a[i].t][j]=max(f[a[i].t][j],f[a[i].s][j-a[i].w]+a[i].v); } } } for(int i=1;i<=B;i++)ans=max(ans,f[L][i]); printf("%d",ans); }