HDU 1712 分组背包

时间：2021-09-07 05:16:23

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4912    Accepted Submission(s): 2651

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
 
 
题目意思：
有n个课程，每个课程在不同天数得到不同价值，总共m天，问m天之内得到最大的价值。
 
思路：
每个课程若在i天完成，就不能在j天完成。以n个课程为组，天数为体积，价值为价值，即得到分组背包的模型。
 
dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[k]]+w[k])//dp[i][j]表示前i组已占用j体积时最大价值。
 
代码：
 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <iostream>

 #include <vector>

 #include <queue>

 #include <cmath>

 #include <set>

 using namespace std;

 #define N 105

 int max(int x,int y){return x>y?x:y;}

 int min(int x,int y){return x<y?x:y;}

 int abs(int x,int y){return x<?-x:x;}

 int a[N][N];

 int n, m;

 int dp[N];

 main()

 {

     int i, j, k;

     while(scanf("%d %d",&n,&m)==){

         if(!n&&!m) break;

         for(i=;i<=n;i++){

             for(j=;j<=m;j++)

                 scanf("%d",&a[i][j]);

         }

         memset(dp,,sizeof(dp));

         for(i=;i<=n;i++){

             for(j=m;j>=;j--){

                 for(k=;k<=j;k++){

                     dp[j]=max(dp[j],dp[j-k]+a[i][k]);

                 }

             }

         }

         printf("%d\n",dp[m]);

     }

 }

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