BZOJ 1649: [Usaco2006 Dec]Cow Roller Coaster (dp，类似01背包)

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 504 Solved: 265
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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
选用第3条，第5条和第6条钢轨

思路：一个类似01dp的背包

dp[i][j]表示长度为i，成本为j的最大有趣指数。

这个想到了。。

然后转移没有推出来。。。

因为潜意识里总觉得循环的变量是要和dp数组的两维一致。。。

实际上根本无关好吗23333

初始化数组dp为-1，dp[0][0]=0

为-1是因为无解输出-1，这样就不用特别处理了。。。

转移方程为：f[i][end[a]]=max{f[i-cost[a][begin[a]]+w[a]} 当f[i-cost[a][begin[a]]可行时

参考了这篇题解：

bzoj1649题解




/* ***********************************************
Author :111qqz
Created Time :2016年04月11日 星期一 16时43分19秒
File Name :code/bzoj/1649.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E4+3;
int n,l,b;
int dp[1005][1005];
struct node
{
    int l,r;
    int w;
    int f;
    int c;

    bool operator < (node b) const
    {
	if (l==b.l) return r<b.r;
	return l<b.l;
    }

}p[N];
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif

	scanf("%d %d %d",&l,&n,&b);
	for ( int i = 1 ; i <= n ; i++)
	{
	    scanf("%d %d %d %d",&p[i].l,&p[i].w,&p[i].f,&p[i].c);
	    p[i].r = p[i].l + p[i].w;
	}

	sort(p+1,p+n+1);

	ms(dp,-1);
	dp[0][0] = 0;
	for ( int i = 1 ; i <= n ; i++)
	{
	    for ( int j = p[i].c ;  j <= b ; j++) 
		if (dp[p[i].l][j-p[i].c]!=-1) dp[p[i].r][j]=max(dp[p[i].r][j],dp[p[i].l][j-p[i].c]+p[i].f);
	}
	int ans = -1 ; //无解输出-1
	for ( int i = 1 ;i <= b ; i++) ans = max(ans,dp[l][i]);
	printf("%d\n",ans);



  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

/* ***********************************************

Author :111qqz

Created Time :2016年04月11日星期一 16时43分19秒

File Name :code/bzoj/1649.cpp

************************************************ */

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <cmath>

#include <cstdlib>

#include <ctime>

#define fst first

#define sec second

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define ms(a,x) memset(a,x,sizeof(a))

typedef long long LL;

#define pi pair < int ,int >

#define MP make_pair

using namespace std;

const double eps = 1E-8;

const int dx4[4]={1,0,0,-1};

const int dy4[4]={0,-1,1,0};

const int inf = 0x3f3f3f3f;

const int N=1E4+3;

int n,l,b;

int dp[1005][1005];

struct node

{

int l,r;

int w;

int f;

int c;

bool operator < (node b) const

{

if (l==b.l) return r<b.r;

return l<b.l;

}

}p[N];

int main()

{

#ifndef ONLINE_JUDGE

freopen("code/in.txt","r",stdin);

#endif

scanf("%d %d %d",&l,&n,&b);

for ( int i = 1 ; i <= n ; i++)

{

scanf("%d %d %d %d",&p[i].l,&p[i].w,&p[i].f,&p[i].c);

p[i].r = p[i].l + p[i].w;

}

sort(p+1,p+n+1);

ms(dp,-1);

dp[0][0] = 0;

for ( int i = 1 ; i <= n ; i++)

{

for ( int j = p[i].c ; j <= b ; j++)

if (dp[p[i].l][j-p[i].c]!=-1) dp[p[i].r][j]=max(dp[p[i].r][j],dp[p[i].l][j-p[i].c]+p[i].f);

}

int ans = -1 ; //无解输出-1

for ( int i = 1 ;i <= b ; i++) ans = max(ans,dp[l][i]);

printf("%d\n",ans);

#ifndef ONLINE_JUDGE

fclose(stdin);

#endif

return 0;

}

作者： CrazyKK

ex-ACMer@hust，researcher@sensetime 查看CrazyKK的所有文章