ACboy needs your help
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
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题解:
分组背包问题:
for 所有的组k
for v=V..0
for 所有的i属于组k
f[v]=max{f[v],f[v-c[i]]+w[i]}
注意循环的顺序,v在i里面,保证每组最多只有一个物品被选
#include<algorithm> #include<iostream> #include<cstdlib> #include<cstdio> using namespace std; int f[110],w[110][110]; int get() { int ans=0,f=1;char ch=getchar(); while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();} while(isdigit(ch)) {ans=ans*10+ch-'0';ch=getchar();} return ans*f; } void init(int m) { fill(f,f+m+1,0); } int main() { int n,m; while(1) { n=get();m=get(); if(n==0&&m==0) return 0; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) w[i][j]=get(); init(m); for(int i=1;i<=n;i++) for(int j=m;j>=0;j--) for(int k=0;k<=j;k++) f[j]=max(f[j],f[j-k]+w[i][k]); printf("%d ",f[m]); } return 0; }