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: [e-mail]     > 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;}