Cow Roller Coaster 典型限定状态DP（或称枚举状态DP）

 

Cow Roller Coaster
Time Limit:2000MS  Memory Limit:65536K
Total Submit:797 Accepted:236

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 W_i (1 ≤ W_i ≤ L). Due to varying terrain, each component i can be only built starting at location X_i (0 ≤ X_i ≤ L - W_i). 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" F_i (1 ≤ F_i ≤ 1,000,000) and a cost C_i (1 ≤ C_i ≤ 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: X_i, W_i, F_i, and C_i.

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

 

Hint
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.

Source
USACO 2006 December Silver

在DP题中 状态的选择往往是至关重要的 要满足能完整描叙一个状态的信息 不遗漏 不多余
而有时候却发现一个状态的目标元素有两个。比如这个题目中典型的可以看出 费用和收益是一个状态的2个目标元素。而一般而言碰到这种情况 我们都会把其中一个元素放到状态中去 是一个状态的目标值确定下来做DP 其实从本质上来说 是一个枚举性质的DP 即枚举一个目标元素的所有可能值 在某个具体的值下面做DP
所以 聪明的朋友一定想到了 我们要尽可能将范围小的元素放入状态中 对
所以此题的状态设置为dp[i][j] 代表修建房子到i地址时使用费用为j的时候的最大收益值
这题的状态转移是分散的 因此最好是对于每个Component做一次转移 呵呵
讲了这么多理论 其实没什么深奥的 背包模型想大家应该都非常熟悉了 其实背包模型不是上面所说的一种特例么？只是前面所说的方法可以应用与2个甚至多个目标元素的情况

 1 #include  < string .h >
 2 #include  < stdio.h >
 3 #include  < algorithm >
 4 using namespace std;
 5
 6 const   int  MAXINT  =   200000000 ;
 7 const   int  N  =   1010 ;
 8 const   int  M  =   10010 ;
 9 int  L, n, cost;
10 int  dp[N][N];  // iµØÖ·, C×Ücost
11 struct Node {  int  st, l, f, c; } p[M];
12
13 bool operator < ( const  Node &  a,  const  Node &  b) { return a.st  <  b.st; }
14
15 #define Max(a, b) ((a)  >  (b) ? (a) : (b))
16
17 int  main() {
18  scanf( " %d %d %d " ,  & L,  & n,  & cost);
19   int  i, j;
20   for (i  =   0 ; i  <  n;  ++ i) {
21   scanf( " %d %d %d %d " ,  & p[i].st,  & p[i].l,  & p[i].f,  & p[i].c);
22  }
23
24  sort(p, p  +  n);
25
26   for (i  =   0 ; i  <=  L;  ++ i)
27    for (j  =   0 ; j  <=  cost;  ++ j)
28    dp[i][j]  =   - MAXINT;
29  dp[ 0 ][ 0 ]  =   0 ;
30
31   for (i  =   0 ; i  <  n; i ++ ) {
32    int &  st  =  p[i].st,  & l  =  p[i].l,  & f  =  p[i].f,  & c  =  p[i].c;
33    for (j  =   0 ; j  <  cost;  ++ j)  if (dp[st][j]  >=   0 ) {
34    dp[st  +  l][j  +  c]  =  Max( dp[st + l][j + c], dp[st][j]  +  f );
35   }
36  }
37
38   int  max  =   - MAXINT;
39   for (i  =   0 ; i  <=  cost;  ++ i) {  // cost 
40    if (dp[L][i]  >  max)
41    max  =  dp[L][i];
42  }
43
44   if (max  ==   - MAXINT) max  =   - 1 ;
45
46  printf( " %d\n " , max);
47
48  return  0 ;
49 }
50