bzoj 1649: [Usaco2006 Dec]Cow Roller Coaster（DP）

1649: [Usaco2006 Dec]Cow Roller Coaster

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 710  Solved: 358

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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

dp[i][j]表示前i米都铺完且费用为j的最大价值

第二维类似于01背包

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define LL long long
typedef struct Res
{
int x, y;
int cost, val;
bool operator < (const Res &b) const
{
if(y<b.y)
return 1;
return 0;
}
}Res;
Res s[10005];
LL dp[1005][1005];
int main(void)
{
LL ans;
int L, n, V, i, j, p;
scanf("%d%d%d", &L, &n, &V);
for(i=1;i<=n;i++)
{
scanf("%d%d%d%d", &s[i].x, &s[i].y, &s[i].val, &s[i].cost);
s[i].y = s[i].x+s[i].y;
}
sort(s+1, s+n+1);
memset(dp, -1, sizeof(dp));
dp[0][0] = 0;
p = 1;
for(i=1;i<=L;i++)
{
while(s[p].y==i && p<=n)
{
for(j=V;j>=s[p].cost;j--)
{
if(dp[s[p].x][j-s[p].cost]==-1)
continue;
dp[i][j] = max(dp[i][j], dp[s[p].x][j-s[p].cost]+s[p].val);
}
p++;
}
}
ans = -1;
for(i=0;i<=V;i++)
{
if(dp[L][i]==-1)
continue;
ans = max(ans, dp[L][i]);
}
printf("%lld\n", ans);
return 0;
}