HDOJ1712 分组背包问题

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7556    Accepted Submission(s): 4178


Problem Description

ACboy has N courses this term, and he plans to spend at most M days on study.Of course,the profit he will gain from different course depending on the days he spend on it.How to arrange the M days for the N courses to maximize the profit?

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers N and M, N is the number of courses, M is the days ACboy has.

Next follow a matrix A[i][j], (1<=i<=N<=100,1<=j<=M<=100).A[i][j] indicates if ACboy spend j days on ith course he will get profit of value A[i][j].

N = 0 and M = 0 ends the input.

 

Output

For each data set, your program should output a line which contains the number of the max profit ACboy will gain.

 

Sample Input

2 2
1 2
1 3
2 2
2 1
2 1
2 3
3 2 1
3 2 1
0 0

 

Sample Output

3
4
6

 

Source

HDU 2007-Spring Programming Contest

 

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问题描述是有N件物品和一个容量为V的背包。第i件物品的费用c[i]，价值w[i]。这些物品被划分为若干组，每组中的物品互相冲突，最多选一件。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。

#include <ctime>
#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;

const int maxn = 105;
int i,j,k,n,m,v;
int val[maxn][maxn];
int dp[maxn];

int main(){
std::ios::sync_with_stdio(false);
while (cin >> n >> m && n+m) {
for (i=1;i<=n; i++)
for (j=1; j<=m; j++) cin >> val[i][j];
memset(dp,0,sizeof(dp));

for (i=1; i<=n; i++) {  //最外层循环分组总数
for (v=m; v>=0;v--) {  //第二层循环总容量
for (k=1;k<=v; k++) //最内层循环每组的种类
dp[v] = max(dp[v],dp[v-k]+val[i][k]);
}
}
cout << dp[m] << endl;
}
return 0;
}