有点类似背包 , 就是那样子搞...

------------------------------------------------------------------------------------

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define rep( i , n ) for( int i = 0 ;  i < n ; ++i )
#define clr( x , c ) memset( x , c , sizeof( x ) )
#define Rep( i , n ) for( int i = 1 ; i <= n ; ++i )
 
using namespace std;
 
const int maxn = int( 1e4 ) + 5;
const int maxl = 1000 + 5;
const int maxc = 1000 + 5;
const int inf = 0x7fffffff;
 
struct data {
int x , l , w , c;
void Read() {
scanf( "%d%d%d%d" , &x , &l , &w , &c );
}
bool operator < ( const data &rhs ) const {
return x < rhs.x || ( x == rhs.x && l < rhs.l );
}
};
 
data A[ maxn ];
int d[ maxl ][ maxc ];
 
int main() {
// freopen( "test.in" , "r" , stdin );
int L , n , MAX;
scanf( "%d%d%d" , &L , &n , &MAX );
MAX++;
rep( i , n ) 
   A[ i ].Read();
   
Rep( i , L )
   rep( j , MAX )
       d[ i ][ j ] = -inf;
rep( i , MAX )
   d[ 0 ][ i ] = 0;
sort( A , A + n );
rep( i , n ) {
data &o = A[ i ];
for( int j = o.c ; j < MAX ; j++ ) {
int &f = d[ o.x + o.l ][ j ];
f = max( f , d[ o.x ][ j - o.c ] + o.w );
}
}
int ans = 0;
rep( i , MAX ) ans = max( ans , d[ L ][ i ] );
ans > 0 ? printf( "%d\n" , ans ) : printf( "-1\n" );
return 0;
}

------------------------------------------------------------------------------------

1649: [Usaco2006 Dec]Cow Roller Coaster

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 440  Solved: 233
[Submit][Status][Discuss]

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.

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).

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.

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

17
选用第3条,第5条和第6条钢轨

HINT

Source

BZOJ 1649: [Usaco2006 Dec]Cow Roller Coaster( dp )的更多相关文章

  1. BZOJ——1649: [Usaco2006 Dec]Cow Roller Coaster

    http://www.lydsy.com/JudgeOnline/problem.php?id=1649 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 7 ...

  2. bzoj 1649: [Usaco2006 Dec]Cow Roller Coaster【dp】

    DAG上的dp 因为本身升序就是拓扑序,所以建出图来直接从1到ndp即可,设f[i][j]为到i花费了j #include<iostream> #include<cstdio> ...

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

    http://www.lydsy.com/JudgeOnline/problem.php?id=1649 又是题解... 设f[i][j]表示费用i长度j得到的最大乐趣 f[i][end[a]]=ma ...

  4. bzoj1649 [Usaco2006 Dec]Cow Roller Coaster

    Description The cows are building a roller coaster! They want your help to design as fun a roller co ...

  5. 【动态规划】bzoj1649 [Usaco2006 Dec]Cow Roller Coaster

    很像背包. 这种在一个数轴上进行操作的题常常需要对区间排序. f[i][j]表示距离到i时,花费为j时的权值之和. f[x[i]+l[i]][j+c[i]]=max{f[x[i]][j]+w[i]}( ...

  6. Bzoj 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 深搜,bitset

    1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 554  Solved: 346[ ...

  7. BZOJ 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐( dfs )

    直接从每个奶牛所在的farm dfs , 然后算一下.. ----------------------------------------------------------------------- ...

  8. BZOJ 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐

    Description The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is graz ...

  9. bzoj 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐【dfs】

    从每个奶牛所在草场dfs,把沿途dfs到的草场的con都+1,最后符合条件的草场就是con==k的,扫一遍统计一下即可 #include<iostream> #include<cst ...

随机推荐

  1. 梯度下降算法的一点认识(Ng第一课)

    昨天开始看Ng教授的机器学习课,发现果然是不错的课程,一口气看到第二课. 第一课 没有什么新知识,就是机器学习的概况吧. 第二课 出现了一些听不太懂的概念.其实这堂课主要就讲了一个算法,梯度下降算法. ...

  2. ocx控件获取使用App的窗口句柄

    在CXxxCtrl文件中 HWND hAppWnd = NULL; if (m_pInPlaceSite != NULL) m_pInPlaceSite->GetWindow(&hApp ...

  3. 64位windows8.1编译openjdk8

    首先下载openjdk8http://hg.openjdk.java.net/jdk8/jdk8/这里用到版本管理工具Mercurial,与git有些类似,不了解的话可以查一下.clone后,里面有个 ...

  4. EL表达式中引用隐式变量

    除了在jsp中9大隐式变量(在前面文章也叫预定义变量)在转化成为servlet后_jspService中可以看到: public void _jspService(final javax.servle ...

  5. BZOJ 1491: [NOI2007]社交网络( floyd )

    floyd...求最短路时顺便求出路径数. 时间复杂度O(N^3) ------------------------------------------------------------------ ...

  6. java Socket 列子 一些参数设置比较全

    http://blog.csdn.net/a19881029/article/details/11596945

  7. BUAA 更大公约数

    题目链接 给一个n*m的矩阵, 删除里面的一行一列, 使得剩下的数的最大公约数最大. 一个格子(x,y), 先预处理出(1,1)到这个格子的内所有数的最大公约数, 同理处理出(1, m), (n, m ...

  8. POJ 1823 Hotel 线段树

    题目链接 线段树的区间合并. 和上一题差不多....第三种操作只需要输出maxx[1]的值就可以. #include <iostream> #include <vector> ...

  9. IOS 特定于设备的开发:使用加速能力“向上定位”

    iPhone提供了3个机载的传感器,用于沿着iPhone的3根相互垂直的轴(左/右(x轴).上/下(y轴)和前/后(z轴))度量加速能力.这些值指示作用于iPhone的力,它们来自重力和用户移动.可以 ...

  10. querySelector $() getElementBy区别

    参考 http://stackoverflow.com/questions/14377590/queryselector-and-queryselectorall-vs-getelementsbycl ...