usaco4.2.3 Job Processing

一 原题

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 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

二 分析

给定N个零件，M台机器，可以贪心（根据机器i结束当前加工的时间t[i]加上生产一个零件的时间s[i]）求出每个零件加工完成的时间T[i]。如果有两道工序A,B，只需要将工序A里最慢的交给工序B里最快的即可，即答案等于max{Ta[i] + Tb[n-1-i]}。

三 代码

运行结果：

USER: Qi Shen [maxkibb3]TASK: jobLANG: C++Compiling...Compile: OKExecuting...   Test 1: TEST OK [0.000 secs, 4188 KB]   Test 2: TEST OK [0.000 secs, 4188 KB]   Test 3: TEST OK [0.000 secs, 4188 KB]   Test 4: TEST OK [0.000 secs, 4188 KB]   Test 5: TEST OK [0.000 secs, 4188 KB]   Test 6: TEST OK [0.000 secs, 4188 KB]   Test 7: TEST OK [0.000 secs, 4188 KB]   Test 8: TEST OK [0.000 secs, 4188 KB]   Test 9: TEST OK [0.000 secs, 4188 KB]   Test 10: TEST OK [0.000 secs, 4188 KB]   Test 11: TEST OK [0.000 secs, 4188 KB]   Test 12: TEST OK [0.000 secs, 4188 KB]All tests OK.
Your program ('job') produced all correct answers!  This is yoursubmission #6 for this problem.  Congratulations!

AC代码：

/*ID:maxkibb3LANG:C++PROB:job*/#include<cstdio>const int MAXN = 1005;const int MAXM = 35;const int INF = 0x7fffffff;int n, m[2];int a[2][MAXM];int t[2][MAXM];int s[2][MAXN];int max(int n1, int n2) {    return (n1 > n2) ? n1 : n2;}void init() {    scanf("%d%d%d", &n, &m[0], &m[1]);    for(int i = 0; i < m[0]; i++) scanf("%d", &a[0][i]);    for(int i = 0; i < m[1]; i++) scanf("%d", &a[1][i]);}void f(int idx) {    for(int i = 0; i < n; i++) {        int min_val = INF, min_idx;        for(int j = 0; j < m[idx]; j++) {            if(t[idx][j] + a[idx][j] < min_val) {                min_val = t[idx][j] + a[idx][j];                min_idx = j;            }        }        s[idx][i] = t[idx][min_idx] = min_val;    }}void solve() {    f(0);    f(1);    int max_val = 0;    for(int i = 0; i < n; i++) {        if(s[0][i] > max_val) max_val = s[0][i];    }    printf("%d ", max_val);    max_val = 0;    for(int i = 0; i < n; i++) {        if(s[0][i] + s[1][n - 1 -i] > max_val) max_val = s[0][i] + s[1][n - 1- i];    }    printf("%d\n", max_val);}int main() {    freopen("job.in", "r", stdin);    freopen("job.out", "w", stdout);    init();    solve();    return 0;}