poj3624 01背包入门 dp+滚动数组

poj3624 01背包 dp+滚动数组

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25458		Accepted: 11455

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7

Sample Output

23
题意：01背包入门题
思路：dp，用二维数组会MLE，由于每个状态只与前一状态有关，故可开滚动数组压缩空间
设放第i个物品的决定（可放可不放）后，已占有容量为j，总价值为dp[i][j],则
dp[i][j]={dp[i-1][j],dp[i-1][w[i]-j]+v[i]} (w[i]-j>0)
边界控制：
dp[i][j]=0 (i==0||j==0)
dp[i][j]=dp[i-1][j] (w[j]-j>0)

下面是我自己的AC代码，遍历j的时候是顺序遍历的，在二维数组或者滚动数组中j可以逆序遍历，也可以顺序遍历，因为dp[i][j]由dp[i-1][j-w[i]]推出，而i在外层循环，递推过程每次先算出dp[i-1][]的所有项，再由dp[i-1][]推出dp[i][]的所有项，在推出dp[i][j]的过程中dp[i-1][j-w[i]]肯定还是dp[i-1][j-w[i]]，因为已经被保存下来了

/* 背包问题  dp+滚动数组  */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<ctype.h>

using namespace std;

const int maxn=14100;
const int INF=(1<<28);
int N,M;
int w[maxn],v[maxn];
int dp[2][maxn];
int main()
{
    scanf("%d%d",&N,&M);
    for(int i=1;i<=N;i++) scanf("%d%d",&w[i],&v[i]);
    memset(dp,0,sizeof(dp));
    for(int i=0;i<=N;i++){
        for(int j=0;j<=M;j++){
            if(i==0||j==0) dp[i%2][j]=0;
            else if(j-w[i]<0) dp[i%2][j]=dp[(i+1)%2][j];
            else dp[i%2][j]=max(dp[(i+1)%2][j],dp[(i+1)%2][j-w[i]]+v[i]);
        }
    }
    printf("%d
",dp[N%2][M]);
    return 0;
}

 下面是逆序遍历的AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<ctype.h>

using namespace std;

const int maxn=14100;
const int INF=(1<<28);
int N,M;
int w[maxn],v[maxn];
int dp[2][maxn];
int main()
{
    scanf("%d%d",&N,&M);
    for(int i=1;i<=N;i++) scanf("%d%d",&w[i],&v[i]);
    memset(dp,0,sizeof(dp));
    for(int i=0;i<=N;i++){
        for(int j=M;j>=0;j--){
            if(i==0||j==0) dp[i%2][j]=0;
            else if(j-w[i]<0) dp[i%2][j]=dp[(i+1)%2][j];
            else dp[i%2][j]=max(dp[(i+1)%2][j],dp[(i+1)%2][j-w[i]]+v[i]);
        }
    }
    printf("%d
",dp[N%2][M]);
    return 0;
}

下面用迭代可以把滚动数组变成一维数组，但变成一维数组后j必须逆序遍历，因为dp[j]=max{dp[j],dp[j-w[i]]+v[i]}，第一个dp[j]表示dp[i][j]，第二个dp[j]和dp][j-w[i]]分别表示dp[i-1][j]和dp[i-1][j-w[i]]，如果顺序遍历，那么第二个dp[j-w[i]]在遍历ｊ过程中会被给成dp[i][j-w[i]]，因为遍历方向是ｊ从小到大，而j比j-w[i]大，所以遍历完dp[i][j-w[i]]后再遍历dp[i][j]时就dp[j-w[i]]就不是dp[i-1][j-w[i]]了，而是dp[i][j-w[i]]！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<ctype.h>

using namespace std;

const int maxn=14100;
const int INF=(1<<28);
int N,M;
int w[maxn],v[maxn];
int dp[maxn];
int main()
{
    scanf("%d%d",&N,&M);
    for(int i=1;i<=N;i++) scanf("%d%d",&w[i],&v[i]);
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=N;i++){
        for(int j=M;j>=0;j--){
            if(j-w[i]>=0) dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
        }
    }
    printf("%d
",dp[M]);
    return 0;
}

 下面是java代码

//2594ms 好慢。。。
import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args){
        Scanner in=new Scanner(System.in);
        int N=in.nextInt();
        int M=in.nextInt();
        int w[]=new int[N+1],v[]=new int[N+1];
        for(int i=1;i<=N;i++){
            w[i]=in.nextInt();
            v[i]=in.nextInt();
        }
        int dp[]=new int[14000];
        dp[0]=0;
        for(int i=1;i<=N;i++){
            for(int j=M;j>=0;j--){
                if(j-w[i]>=0) dp[j]=max(dp[j-w[i]]+v[i],dp[j]);
            }
        }
        System.out.println(dp[M]);
    }
    public static int max(int a,int b){
        return a>b?a:b;
    }
}