USACO 4.2 job processing

A factory is running a production line that requires two operations to be performed on each job: first operation "A" then operation "B". Only a certain number of machines are capable of performing each operation.

Figure 1 shows the organization of the production line that works as follows. A type "A" machine takes a job from the input container, performs operation "A" and puts the job into the intermediate container.
A type "B" machine takes a job from the intermediate container, performs operation "B" and puts the job into the output container. All machines can work in parallel and independently of each other, and the size of each container is unlimited. The machines
have different performance characteristics, a given machine requires a given processing time for its operation.

Give the earliest time operation "A" can be completed for all N jobs provided that the jobs are available at time 0. Compute the minimal amount of time that is necessary to perform both operations (successively,
of course) on all N jobs.

PROGRAM NAME: jobINPUT FORMAT

Line 1:	Three space-separated integers: N, the number of jobs (1<=N<=1000). M1, the number of type "A" machines (1<=M1<=30) M2, the number of type "B" machines (1<=M2<=30)
Line 2..etc:	M1 integers that are the job processing times of each type "A" machine (1..20) followed by M2 integers, the job processing times of each type "B" machine (1..20).

SAMPLE INPUT (file job.in)

5 2 31 1 3 1 4

OUTPUT FORMAT

A single line containing two integers: the minimum time to perform all "A" tasks and the minimum time to perform all "B" tasks (which require "A" tasks, of course).

SAMPLE OUTPUT (file job.out)

3 5

關於求A的最短時間，最先想到的是DP，而總時間的最小值該怎麼求一直沒有好的思路，感覺這個問題和4.1的fence rail 有些相似，最後只好搜尋加剪枝，前6個點能過，第7個點逾時，後來聽ZZY說這個可以用貪心，再上網看了些解題報告，終於豁然開朗。在此推薦一個解釋得比較清晰的部落格：http://magicalcode.blogbus.com/logs/37193487.html

下面是My Code：

#include<cstdio>#include<algorithm>#define oo 1000000using namespace std;FILE *in,*out;int n,a,b,ta[30],tb[30],cost[30],tableA[10001],tableB[10001];void greedy(int *arr,int len,int *rec);bool cmp(const int &a,const int & b);int main(){    in=fopen("job.in","r");    out=fopen("job.out","w");    fscanf(in,"%d%d%d",&n,&a,&b);    for(int i=0;i<a;i++) fscanf(in,"%d",&ta[i]);    for(int i=0;i<b;i++) fscanf(in,"%d",&tb[i]);    greedy(ta,a,tableA);    greedy(tb,b,tableB);    int max=0;    sort(tableA,tableA+n);    fprintf(out,"%d ",tableA[n-1]);    sort(tableB,tableB+n,cmp);    for(int i=0;i<n;i++) if(max<(tableA[i]+tableB[i])) max=tableA[i]+tableB[i];    fprintf(out,"%d\n",max);    fclose(in);    fclose(out);    return 0;}void greedy(int *arr,int len,int *rec){     for(int i=0;i<len;i++) cost[i]=0;     for(int i=0;i<n;i++)    {         int min=oo;         int index;         for(int j=0;j<len;j++)         {              if(min>cost[j]+arr[j])              {                   min=cost[j]+arr[j];                   index=j;              }         }         cost[index]=min;         rec[i]=min;    }}bool cmp(const int &a,const int & b){     return a>b;}