四、学算法啦啦啦——dp，并查集等 [Cloned] E - 01背包

原题：

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.

题意：

01背包问题的原题。

题解：

01背包原题，要做出的决定就是这个物品买还是不买，状态方程是

for(i=0; i<n; i++)

 for(j=m; j>=w[i]; j--) 

dp[j]=max(dp[j],dp[j-w[i]]+v[i]); 

别的就没啥了，原型题

代码：AC

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=12880+5;
int v[maxn];
int w[maxn];
int dp[maxn];
int max(int x,int y)
{
	return x>y?x:y;
}
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
		int i,j;
        for(i=0; i<n; i++)
            cin>>w[i]>>v[i];
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
        {
            for(j=m; j>=w[i]; j--)
            {
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        cout<<dp[m]<<endl;
 
    }
    return 0;
}