poj3624 01背包 dp+滚动数组
| 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 Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 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: Wi and Di
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; } }