POJ 3624 Charm Bracelet

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

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

分析

典型的01背包问题，求n个物品放入背包的最大价值，一个物品的重量为w[i]，价值为v[i]，放入与不放入二选一。
递推式：m[i][j] = max(m[i-1][j], max[i - w[i]][j] + v[i])。
在此处不能用二维数组，会超内存限制。

实现

程序实现如下，要说的是优化运行时间的过程，用scanf/printf代替cin/cout节省 50ms , 状态转移方程由：

int temp = dp[j - w[i]] + d[i];dp[j] = max(dp[j], temp);

换为：

int temp = dp[j - w[i]] + d[i];if (dp[j] < temp) {  dp[j] = temp;}

节省了 110ms 左右，函数调用真费时间。。

#include <iostream>#define N 3402#define M 12881int dp[M]={0};int w[N];int d[N];int main(){//  freopen("in.txt", "r", stdin);    int n, m;    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--) {            //每放一件物品求出重量不大于w[j]时的最大价值            int temp = dp[j - w[i]] + d[i];            if (dp[j] < temp) {                dp[j] = temp;            }        }    }    printf("%d\n", dp[m]);    return 0;}