[POJ_3624] 0-1背包 题解

本人半路出家   算法从未规范的学习过，前几天在学习DP的时候，同伴惊呼：你竟然不知道背包，此乃dp基础...... 

闲来无事，学学背包也好~

废话少说，题目如下：

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19607 Accepted: 8911

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 61 42 63 122 7

Sample Output

23

        题意：现在有N颗宝石，每颗宝石都有自己的重量W以及自己的价值D，现在我们来串手链，已知手链所能承受的最大重量为M，求得在手链能承受的最大重量之下，手链的最大价值。此题是典型的0-1背包。

       给出状态转移方程:  dp[j]=max(dp[j],   dp[j - w[i]]+D[i]);

因为手链的最大价值是不确定的，因此，dp的赋值为0

现给出AC代码

#include <iostream>#include<string.h>#include<cstdio>using namespace std;const int MAX = 3500;const int Max = 13000;int dp[13000];int M[3500];int D[13000];int N,M1;int main(){    int ans = -1;//赋值为负数    memset(dp,0,sizeof(dp));    scanf("%d%d",&N,&M1);    for(int i = 1; i <= N; i++)    {        scanf("%d%d",&M[i],&D[i]);    }    for(int i = 1; i <= N; i++)    {        for(int j = M1; j >= M[i]; j--)        {            dp[j]=max(dp[j],dp[j-M[i]]+D[i]);            ans = max(ans,dp[j]);        }    }   printf("%d\n",ans);    return 0;}