HDU 1712 ACboy needs your help(DP)
2014-07-28 17:03
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[align=left]Problem Description[/align]
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?
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each data set, your program should output a line which contains the number of the max profit ACboy will gain.
[align=left]Sample Input[/align]
[align=left]Sample Output[/align]
题意:n个课程,最多有m天可以用,每个课程花费不同的天数得到的收益不同,第i个课程花费j天来学那么收益为a[i][j],问如何安排收益最大。
首先明确每个课程只会上一次。设dp[i][j]为前i个课程花费j天的最大收益,枚举当前i课程的所有需要的天数k,并选择学还是不学对应的a[i][k]。
于是有 dp[i][j] = max( dp[i][j] , dp[i-1][j] , dp[i-1][j-k]+a[i][k] ) (1<=k<=j) .
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?
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each data set, your program should output a line which contains the number of the max profit ACboy will gain.
[align=left]Sample Input[/align]
2 2 1 2 1 3 2 2 2 1 2 1 2 3 3 2 1 3 2 1 0 0
[align=left]Sample Output[/align]
3 4 6
题意:n个课程,最多有m天可以用,每个课程花费不同的天数得到的收益不同,第i个课程花费j天来学那么收益为a[i][j],问如何安排收益最大。
首先明确每个课程只会上一次。设dp[i][j]为前i个课程花费j天的最大收益,枚举当前i课程的所有需要的天数k,并选择学还是不学对应的a[i][k]。
于是有 dp[i][j] = max( dp[i][j] , dp[i-1][j] , dp[i-1][j-k]+a[i][k] ) (1<=k<=j) .
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; typedef long long LL; const int MAX=0x3f3f3f3f; int a[105][105] , dp[105][105],n,m; int Max(int a,int b,int c) { return max( max(a,b) , c ); } int main() { while (~scanf("%d%d",&n,&m) && n+m) { for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&a[i][j]); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) for(int k=1;k<=j;k++) dp[i][j] = Max( dp[i][j] ,dp[i-1][j] ,dp[i-1][ j-k ]+a[i][k] ); printf("%d\n",dp [m]); } return 0; }
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