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：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total 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 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
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;
}