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http://poj.org/problem?id=3624
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 29444 | Accepted: 13198 |
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
这个是简单的01背包 从我开始接触到背包到现在我还是不懂背包的原理
开始背包的旅程
正常的背包是
i 1....n
j m....0
dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);
但是这个范围较大
如果用二维的会超内存
可以转化成一维的
dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<iostream> using namespace std; #define N 3500 int dp[13000]; int main() { int n,m,w[N],v[N]; while(scanf("%d %d",&n,&m)!=EOF) { 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]) dp[j]=max(dp[j],dp[j-w[i]]+v[i]); } } printf("%d\n",dp[m]); } return 0; }
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原文地址：http://www.cnblogs.com/linliu/p/5037108.html
