题目链接:传送门 题面:
ACboy needs your help
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
题意:
你有天要学门课,表示第门课花了天你可以获得的价值,求如何选择课程能使得你获得的价值最大。
分组背包
一组物品就是每一门课的不同价值
只能取一次
先枚举组数
再枚举体积
再枚举每组中的物品
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <vector>
#include <iomanip>
#define
#define
#define
using namespace std;
int f[A], n, m, a[B][B];
int main() {
while (scanf("%d%d", &n, &m)) {
if (!n and !m) return 0;
memset(f, 0, sizeof f);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
scanf("%d", &a[i][j]);
for (int i = 1; i <= n; i++)
for (int j = m; j >= 0; j--)
for (int k = 1; k <= m; k++)
f[j] = max(f[j], f[j - k] + a[i][k]);
printf("%d\n", f[m]);
}
}