D 分组背包

标签：des   style   blog   color   os   2014   

<span style="color:#3333ff;">/*
——————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————

	copyright    :    Grant Yuan
	time         :    2014.7.18
	algorithm    :    分组背包
	
————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————————

D - 分组背包 基础
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
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*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

int n,m;
int a[101][101];
int dp[101];

int main()
{
	while(1){
		cin>>n>>m;
		if(n==0&&m==0)
			 break;
	    for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++)
			  cin>>a[i][j];
	    memset(dp,0,sizeof(dp));

	    for(int i=1;i<=n;i++)
			for(int j=m;j>=1;j--)
			  for(int k=1;k<=j;k++){
			  dp[j]=max(dp[j],dp[j-k]+a[i][k]);
			}
	    cout<<dp[m]<<endl;
	    }
	return 0;
}
</span>

D 分组背包,布布扣,bubuko.com

D 分组背包

标签：des   style   blog   color   os   2014   

原文地址：http://blog.csdn.net/yuanchang_best/article/details/37936519