POJ 3624 Charm Bracelet（01背包问题）

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16295 Accepted: 7403

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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 61 42 63 122 7

Sample Output

23

Source

USACO 2007 December Silver

二维数组求解

/*二维数组求解但是此题Memory限制  AC不了  只能用一维数组存储写二维数组  仅供学习*/#include <iostream>using namespace std;const int N = 3405;const int M = 12885;#define max(a, b) (a) > (b) ? (a) : (b)int dp[N][M];int W[N];int P[N];int main(){int n, m;scanf("%d %d", &n, &m);for(int i = 1; i <= n; i++) scanf("%d %d", &W[i], &P[i]);//注意，i==1或者j==1时，会用到dp边界下标为0的元素，由于已经初始化这些元素为0，保证了程序的正确运行for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++) {if(W[i] <= j)dp[i][j] = max(dp[i-1][j-W[i]] + P[i], dp[i-1][j]);elsedp[i][j] = dp[i-1][j];}printf("%d\n", dp[n][m]);return 0;}

一维数组求解

//一维数组求解#include<cstdio>#include<cstring>#define N 3500#define M 13000#define max(a, b) (a) > (b) ? (a) : (b)int n, m;int W[N], P[N];int dp[M];int main(){while(scanf("%d %d", &n, &m) != EOF){for(int i = 1; i <= n; i++){scanf("%d", &W[i]);scanf("%d", &P[i]);}memset(dp, 0 , sizeof(dp));for(int i = 1; i <= n; i++)for(int j = m; j >= W[i]; j--)dp[j] = max(dp[j], dp[j - W[i]] + P[i]);printf("%d\n", dp[m]);}return 0;}