0-1背包问题，poj 3624 Charm Bracelet动态规划-解题报告，增加最优路径构建

1）0-1背包问题和 零碎背包问题是不同的，前者只能用动态规划来做， 后者可以用贪心算法。

2）动态规划的核心是 “有多个重叠子问题”，“自底向上”解决问题。

3) 0-1背包问题 ，W为最大重量，n为物体个数，求最大的价值Value，可在O(nW)的时间复杂度内解算出来。

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Charm Bracelet

Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 10014
Accepted: 4498

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 theN (1 ≤N ≤ 3,402) available charms. Each charm i in the supplied list has a weightW_i (1 ≤W_i ≤ 400), a 'desirability' factorD_i (1 ≤D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more thanM (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 andD_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

Source

USACO 2007 December Silver

#include <stdio.h>int N,M;int W[3410],D[3410];int f[12882];int rout[12882]; //寻根，帮助重建 int main(){   scanf("%d%d",&N,&M);   for(int i=1;i<=N;i++)   {       scanf("%d%d",&W[i],&D[i]);   }     for(int i=1;i<=M;i++)   {       f[i]=0;      rout[i]=-1;  }       for(int i=1;i<=N;i++)   //注意不是从W[i]到M，而是从M到W[i],因为这次是基于上次的基础（后者），而不是重复算两次当下（前者）       for(int j=M;j>=W[i];j--)      {          if(f[j-W[i]] +D[i]>f[j])          {                    f[j]=f[j-W[i]]+D[i];                    rout[j]=W[i];                    }          }       int ma=-1;   int pos=-1;   for(int i=1;i<=M;i++)      if(f[i]>ma)      {       ma=f[i];       pos=i;      }         printf("%d\n",ma);  int j=pos;   while(rout[j]!=-1 && j>0)   {       printf("%d\n",rout[j]);       j=j-rout[j];   }          scanf("%d",&N);    return 0;}