ＰＯＪ　３２５７　Cow Roller Coaster　二维背包

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.

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

输入样例#1：复制

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

输出样例#1：复制

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.

算法分析

题意：

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)．

分析：

二维背包问题

DP[i][j]表示在i位置花费为j时最大有趣值。

初始化：DP[i][j]为-1

DP[0][0]=0

状态转移方程：dp[a[i].e][j]=max(dp[a[i].e][j],dp[a[i].s][j-a[i].c]+a[i].f);

状态转移方程条件：if(dp[a[i].s][j-a[i].c]>=0) //判断起点是否可以走得通

代码实现

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
const int M=10010;
struct node
{int s,e,f,c;
};
bool cmp(const node &a,const node &b)
{if(a.s==b.s)return a.e<b.e;else return a.s<b.s;
}
node a[10010];
int dp[1010][1010];
int main()
{int l,n,b;while(scanf("%d%d%d",&l,&n,&b)!=EOF){for(int i=1;i<=n;i++){int t;scanf("%d%d%d%d",&a[i].s,&t,&a[i].f,&a[i].c);a[i].e=a[i].s+t;}sort(a+1,a+n+1,cmp);memset(dp,-1,sizeof(dp));dp[0][0]=0;for(int i=1;i<=n;i++)for(int j=b;j>=a[i].c;j--){if(dp[a[i].s][j-a[i].c]>=0)  //判断起点是否可以走得通dp[a[i].e][j]=max(dp[a[i].e][j],dp[a[i].s][j-a[i].c]+a[i].f);}int ans=-1;for(int i=1;i<=b;i++)ans=max(ans,dp[l][i]) ;printf("%d\n",ans);}return 0;
}