hdu 1712 ACboy needs your help 分组背包

ACboy needs your help

Time Limit: ¹⁰⁰⁰⁄₁₀₀₀ MS (Java/Others) Memory Limit: ³²⁷⁶⁸⁄₃₂₇₆₈ K (Java/Others) Total Submission(s): 2403 Accepted Submission(s): 1223

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

分组背包，唉，开始写还是卡了。。卡在第三层循环有木有！j表示花费，所以呢，只能从1循环到v，不能超过去了。也就是说，对于每一组中的物品，要首先看一下这个物品的花费是不是比v小，然后再考虑要不要放进去。还是看的别人的代码才明白的。。次嗷……以后再也不敢了。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int a[120][120];
int main(void){
#ifndef ONLINE_JUDGE
  freopen("1712.in", "r", stdin);
#endif
  int n, m, f[200];
  while (~scanf("%d%d", &n, &m)){
    memset(f, 0, sizeof(f));
    if (m+n==0) break;
    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 v = m; v >= 0; --v){
        for (int j = 1; j <= v; ++j){
          f[v] = max(f[v], f[v-j] + a[i][j]);
        }
      }
    }
    printf("%d\n", f[m]);
  }

  return 0;
}

理解，关键是理解！不能盲目照着人家的板子抄。。。