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.
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。
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。
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]);
}
}