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| 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]} (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)
/* 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\n",dp[N%2][M]); return 0; }
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原文地址：http://www.cnblogs.com/--560/p/4340764.html
