HDU_1712_背包问题

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 21 21 32 22 12 12 33 2 13 2 10 0 

 

Sample Output

 346 

 

题意就是告诉你每门课程话费一定天数，所能获得的知识。

让你列一个计划，让你所能获得的知识最大

dp[i][j]的意义是学习前i们课程话费j天所能获得的最大知识

我们首先应该明确状态转移方程

dp[i+1][j] = max(dp[i][j]+dp[i][j-k]+A[i][k]||k<=j)

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define MAX_N 105#define MAX_M 105int n;//课程数目int m;//总天数//dp[i+1][j]表示学习前i门课程消耗j天所获得最大价值/**    //状态转移方程    dp[i+1][j]=max(dp[i][j],dp[i][j-k]+A[i][k]|k<=j)*/int dp[MAX_N][MAX_M];//A[i][j]代表花j天在第i门课上所获得的价值int A[MAX_N][MAX_M];int main(){    //freopen("背包D.txt","r",stdin);    while(~scanf("%d %d",&n,&m))    {        if(n==m&&m==0)        {            break;        }        for(int i=0;i<n;i++)        {            for(int j=1;j<=m;j++)            {                scanf("%d",&A[i][j]);            }        }        for(int i=0;i<n;i++)        {            A[i][0]=0;        }        void solve();        solve();    }}void solve(){    memset(dp,0,sizeof(dp));    for(int i=0;i<n;i++)    {        for(int j=0;j<=m;j++)        {

<span style="white-space:pre"></span>    //对所能学习的天数进行枚举            for(int k=0;k<=j;k++)            {                //dp[i][j]代表学习前i门课程所占j天所获得的最大价值                dp[i+1][j]=max(dp[i+1][j],dp[i][j-k]+A[i][k]);            }        }    }    printf("%d\n",dp[n][m]);}