【BZOJ】1649: [Usaco2006 Dec]Cow Roller Coaster(dp)




f[i][end[a]]=max{f[i-cost[a][begin[a]]+w[a]} 当f[i-cost[a][begin[a]]可行时



#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=10005, M=1005;
struct dat { int x, w, f, c; } a[N];
inline const bool cmp(const dat &a, const dat &b) { return a.x<b.x; }
int f[M][M], n, l, m;
int main() {
	read(l); read(n); read(m);
	for1(i, 1, n) read(a[i].x), read(a[i].w), read(a[i].f), read(a[i].c);
	sort(a+1, a+1+n, cmp);
	int ans=-1;
	CC(f, -1); f[0][0]=0;
	for1(i, 1, n) {
		int x=a[i].x, c=a[i].c, e=x+a[i].w, ff=a[i].f;
		for1(j, c, m) if(f[j-c][x]!=-1) f[j][e]=max(f[j][e], f[j-c][x]+ff);
	for1(i, 1, m) ans=max(f[i][l], ans);
	return 0;





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).


* 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.



* 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

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




posted @ 2014-09-07 12:11  iwtwiioi  阅读(530)  评论(0编辑  收藏  举报