POJ 3624 Charm Bracelet【裸01背包和关于贪心解 01 背包问题的思考】
原题链接：http://poj.org/problem?id=3624
我的链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=20437#problem/A欢迎来水
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14963 | Accepted: 6840 |
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
Source
题意：第一行的两个整数n 和 m 分别表示可选择放入的物品有n个，背包体积为m，
接下来 n行每行的前一个整数代表当前物品放入背包的体积和产生的价值。
每种物品只可以放一次，求放入物品总体积不超过背包总体积情况下可以产生的最大
价值。
算法：裸01背包。。。一切重头开始吧
状态转移方程：
for i=1..N
ZeroOnePack(c[i],w[i]);
for v=V..cost
f[v]=max{f[v],f[v-cost]+weight}
//Accepted 216 KB 297 ms C++ 343 B 2013-03-13 22:00:10
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 12880 + 10;
int dp[maxn];
int main()
{
int n,m;
int w,d;
scanf("%d%d", &n, &m);
for(int i =1; i <= n; i++)
{
scanf("%d%d", &w, &d);
for(int j = m; j >=w; j--)
{
dp[j] = max(dp[j],dp[j-w]+d);
}
}
printf("%d\n", dp[m]);
return 0;
}