poj 4131:Charm Bracelet 动规

题目

总时间限制: 1000ms 内存限制: 65536kB
描述
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 iin 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.

输入
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
输出
Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
样例输入
4 6
1 4
2 6
3 12
2 7
样例输出
23

思路

动规题目，首先找到子问题。要解决的问题是N个物品，限重M，求价值最大。可以考虑某一个物品是否选中的最大值，即

1. 该物品选中：该物品的价值加上剩下N-1个物品的最大价值
2. 该物品没选中：剩下N-1个物品的最大价值。

其中，第一种需要考虑剩下N-1种物品的限重也是减小的。

递推公式：$V\_MAX[i,j]=MAX(V\_MAX[i?1,j],V\_MAX[i?1,j?w[i]]+d[i])$ ，其中$V\_MAX[i,j]$是第一i物品限重为j时的最大值，$w[i]$是第i个物品的质量 ，$d[i]$是第i个物品的价值。$V\_MAX[i?1,j]$是不选第i个物品，价值为前$i?1$个限重为$j$时的价值；$V\_MAX[i?1,j?w[i]]+d[i]$是选第$i$个物品限重为$j$的价值，这里是减是因为限重是已经固定的，所以只能是调整前$i?1$个物品的质量，如果是加的话，限重就不是$j$了

该题由于数组太大，内存限制会超，故使用滚动数组记录，只保留一个一维数组来记录

#include <stdio.h>
#include<algorithm>
int N,M;
int w[3405];
int d[3405];
int lo[12880];
using namespace std;int main()
{scanf("%d%d",&N,&M);for(int i=0;i<N;i++){scanf("%d%d",&(w[i]),&(d[i]));}for(int i=0;i<N;i++){for(int j=M;j>=w[i];j--){lo[j] = max(lo[j],lo[j-w[i]]+d[i]);}}printf("%d",lo[M]);return 0;
}