POJ 3624 Charm Bracelet(가방 문제 01)

2500 단어 dp
Charm Bracelet
Time Limit: 1000MS
 
Memory Limit: 65536K
Total 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 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
USACO 2007 December Silver
2차원수 그룹 구해
/*
      
    Memory    AC             
           
*/
#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]);


	//  ,i==1  j==1 ,   dp     0   ,            0,          
	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]);
			else
				dp[i][j] = dp[i-1][j];
		}

		printf("%d
", dp[n][m]); return 0; }

1차원 그룹 구해
//      
#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
", dp[m]); } return 0; }

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