ACboy needs your help(DP)

ACboy needs your help

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4542    Accepted Submission(s): 2431

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.

 

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: 题目给出n个课程和m天,问如何安排m天的课程使得最后的收益达到最大值.题解: 简单DP,以dp[i]表示当前安排i天可以获得的最大收益,则可以得出状态转移方程dp[k + j] = max(dp[k + j], dp[k] + a[i][j]); 表示当前的课程安排j天的最优解. 具体看下代码吧!AC代码:
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-9using namespace std;typedef long long ll;typedef pair<int,int>P;const int M = 200;const int INF = 0x3f3f3f3f;const int mod = 10000007;int maxn[M][M],dp[M],tp[M];int main(){    //freopen("in","r",stdin);    int n,m;    while(~scanf("%d %d",&n,&m) && (n | m)){        memset(dp,0,sizeof(dp));        for(int i = 1; i <= n; i++){            for(int j = 1; j <= m; j++){                scanf("%d",&maxn[i][j]);            }        }        for(int i = 1; i <= m; i++) dp[i] = maxn[1][i]; //初始化        for(int i = 2; i <= n; i++){            for(int k = 1; k <= m; k++) tp[k] = dp[k]; //防止当前的状态会覆盖之前的值,所以先把最优值保存在tp数组中;            for(int j = 1; j <= m; j++){                for(int k = 0; k + j <= m; k++){                    tp[k + j] = max(tp[k + j],dp[k] + maxn[i][j]);                }            }            for(int k = 1; k <= m; k++) dp[k] = max(tp[k],dp[k]);        }        int Max = 1;        for(int i = 2; i <= m; i++)            if(dp[Max] < dp[i]) Max = i;        printf("%d\n",dp[Max]);    }    return 0;}