• hdu1712ACboy needs your help( dp )


    题目大意就是给出 N门课, 然后给出每门课花m天的收益,问在天数一定的情况下最大收益

    ans[x][r] 表示前x节课在r天的情况下的最佳情况

    for ( int i= 0, r)

    ans[x][r] = max( ans[x][r] , ans[x+1][r-i] + cl[x][i])

    用了for和dfs两种方法写

    题目:

    ACboy needs your help

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 3368    Accepted Submission(s): 1742


    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
     
    Recommend
    lcy
     
    代码:
     1 //Sat Jan 11 18:55:27 2014
     2 //Author:Minshik
     3 #include <iostream>
     4 #include <algorithm>
     5 #include <cstring>
     6 #include <cstdio>
     7 #include <string>
     8 #include <vector>
     9 #include <set>
    10 #include <stack>
    11 #include <queue>
    12 #include <deque>
    13 #include <memory.h>
    14 #include <cctype>
    15 #include <cmath>
    16 #include <list>
    17 using namespace std;
    18 
    19 int TIAN,KE;
    20 int cl[105][105];
    21 int ans[105][110];
    22 
    23 #define REP(i,N) for(int i=0;i<N;i++)
    24 
    25 int dfs(int x,int r)
    26 {
    27     if(ans[x][r]) return ans[x][r];
    28 
    29     if(x==KE)return 0;
    30 
    31     int &d = ans[x][r];
    32 
    33     d = dfs(x+1,r);
    34 
    35     for(int i=1;i<=r;i++)
    36     {
    37         d = max(d,dfs(x+1,r-i)+cl[x][i-1]);
    38     }
    39     return d;
    40 
    41 }
    42 int main()
    43 {
    44 
    45     while(cin>>KE>>TIAN && KE+TIAN!=0)
    46     {
    47         REP(i,KE)REP(j,TIAN)
    48                 cin>>cl[i][j];
    49 
    50 
    51 
    52         for(int i=1;i<=TIAN;i++)
    53         {
    54             for(int j=1;j<=KE;j++)
    55             {
    56                  ans[j][i] = ans[j-1][i];
    57                 for(int k=1;k<=i;k++)
    58                 {
    59                      ans[j][i] = max (ans[j][i], ans[j-1][i-k] +cl[j-1][k-1] );
    60                 }
    61 
    62             }
    63         }
    64 
    65 
    66 
    67 
    68         cout<<ans[KE][TIAN]<<endl;
    69 
    70 
    71     }
    72 
    73 
    74     return 0;
    75 }
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  • 原文地址:https://www.cnblogs.com/doubleshik/p/3515499.html
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