hdu1712---ACboy needs your help（dp，分组背包）

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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显然的分组背包问题

dp[i][j] 表示 前i门课共花的时间<=j时，可以获得的最多的价值

转移即可

当然状态可以优化到一维

循环顺序不能错

/*************************************************************************
    > File Name: hdu1712.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年05月10日 星期日 13时52分46秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int val[110][110];
int dp[110];

int main() {
    int n, m;
    while (~scanf("%d%d", &n, &m)) {
        if (!n && !m) {
            break;
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                scanf("%d", &val[i][j]);
            }
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; ++i) {
            for (int j = m; j >= 0; --j) {
                for (int k = 1; k <= m; ++k) {
                    if (j >= k && dp[j - k] != -inf) {
                        dp[j] = max(dp[j], dp[j - k] + val[i][k]);
                    }
                }
            }
        }
        printf("%d\n", dp[m]);
    }
    return 0;
}