USACO: Job Processing

Job Processing
IOI'96

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: job

INPUT 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 3
1 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

思路：

1、独立考虑完成A工序和B工序，简单的贪心策略，每次加入一个新job，选择可以最快完成的机器来加工这个job

2、合并A工序和B工序，这是一个瓶颈问题，为使最长时间最短，记得高中时候学过柯西不等式什么的，多维的柯西不等式就可以应用到这里，两个序列一个升序一个降序，对应想加之后得到的和的最大值最小。

 

/* ID: geterns1 PROG: job LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; const int infi = 1000000; const int size = 1000; const int mach = 30; // int jobs, cntA, timA[mach], cntB, timB[mach]; //X_able[i]: the next free time for machine i of type X int A_able[mach], B_able[mach], A_done[size], B_done[size]; int main(){ freopen("job.in", "r", stdin); freopen("job.out", "w", stdout); // scanf("%d %d %d/n", &jobs, &cntA, &cntB); for (int i = 0; i < cntA; i ++) scanf("%d", timA + i); for (int i = 0; i < cntB; i ++) scanf("%d", timB + i); // int shortest, index; memset(A_able, 0, sizeof(A_able)); memset(B_able, 0, sizeof(B_able)); for (int i = 0; i < jobs; i ++){ // shortest = infi; for (int j = 0; j < cntA; j ++){ if (A_able[j] + timA[j] < shortest){ shortest = A_able[j] + timA[j]; index = j; } } A_done[i] = shortest; A_able[index] += timA[index]; // shortest = infi; for (int j = 0; j < cntB; j ++){ if (B_able[j] + timB[j] < shortest){ shortest = B_able[j] + timB[j]; index = j; } } B_done[i] = shortest; B_able[index] += timB[index]; } // int needed = 0; for (int i = 0; i < jobs; i ++) if (A_done[i] > needed) needed = A_done[i]; printf("%d ", needed); needed = 0; for (int i = 0; i < jobs; i ++) if (A_done[i] + B_done[jobs - 1 - i] > needed) needed = A_done[i] + B_done[jobs - 1 - i]; printf("%d/n", needed); fclose(stdin); fclose(stdout); return 0; }