POJ3624Charm Bracelet（01背包）

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15375		Accepted: 7013

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
此题用二维数组很可能超内存，需要用一维数组求解，因为定义个全局变量n有在主函数里面定义个局部变量n，以为是逻辑错误，调试了近两个小时。。。
其实这个程序w[i]和v[i]数组都可以用两个变量代替，详见刘汝佳算法经典入门。。。

//date 2013.4.9 POJ 3624	Accepted	300K	282MS	C++	897B
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 3500;
const int maxm = 13000;
int w[maxn], v[maxn];
int f[maxm];//数组开小runtime error！
int weight;
int n;
using namespace std;

void DP() {
for(int i = 1; i <= n; i++) {
for(int j = weight; j >= w[i]; j--) {
f[j] = max(f[j], f[j-w[i]]+v[i]);
}
}
}

void init() {
memset(f, 0, sizeof(f));
memset(v, 0, sizeof(v));
memset(w, 0, sizeof(w));
}

int main()
{
//int n;   注意局部变量可以隐藏全局变量，一个多小时的代价！
while(scanf("%d%d", &n, &weight) != EOF) {
for(int i = 1; i <= n; i++) {
cin >> w[i];
cin >> v[i];
}
DP();
cout <<  f[weight] << endl;
init();
}
return 0;
}
/*********************
TEST：
4 6
1 4
2 6
3 12
2 7
****************/