【动态规划】Batch Scheduling

Description
There is a sequence of N jobs to be processed on one machine. The jobs are numbered from 1 to N,
so that the sequence is 1,2,..., N. The sequence of jobs must be partitioned into one or more
batches, where each batch consists of consecutive jobs in the sequence. The processing starts at
time 0. The batches are handled one by one starting from the first batch as follows. If a batch b
contains jobs with smaller numbers than batch c, then batch b is handled before batch c. The jobs
in a batch are processed successively on the machine. Immediately after all the jobs in a batch
are processed, the machine outputs the results of all the jobs in that batch. The output time of
a job j is the time when the batch containing j finishes.

A setup time S is needed to set up the machine for each batch. For each job i, we know its cost
factor Fi and the time Ti required to process it. If a batch contains the jobs x, x+1,... , x+k,
and starts at time t, then the output time of every job in that batch is t + S + (Tx + Tx+1 + ...
+ Tx+k). Note that the machine outputs the results of all jobs in a batch at the same time. If the
output time of job i is Oi, its cost is Oi * Fi. For example, assume that there are 5 jobs, the
setup time S = 1, (T1, T2, T3, T4, T5) = (1, 3, 4, 2, 1), and (F1, F2, F3, F4, F5) = (3, 2, 3, 3,
4). If the jobs are partitioned into three batches {1, 2}, {3}, {4, 5}, then the output times (O1,
O2, O3, O4, O5) = (5, 5, 10, 14, 14) and the costs of the jobs are (15, 10, 30, 42, 56), respectively.
The total cost for a partitioning is the sum of the costs of all jobs. The total cost for the example
partitioning above is 153.

You are to write a program which, given the batch setup time and a sequence of jobs with their
processing times and cost factors, computes the minimum possible total cost.

Input
Your program reads from standard input. The first line contains the number of jobs N, 1 <= N <=
10000. The second line contains the batch setup time S which is an integer, 0 <= S <= 50. The
following N lines contain information about the jobs 1, 2,..., N in that order as follows. First
on each of these lines is an integer Ti, 1 <= Ti <= 100, the processing time of the job. Following
that, there is an integer Fi, 1 <= Fi <= 100, the cost factor of the job.

Output
Your program writes to standard output. The output contains one line, which contains one integer:
the minimum possible total cost.

Sample Input
5
1
1 3
3 2
4 3
2 3
1 4

Sample Output
153

不难想出一个比较简单的方程：
f[i] = min{f[j] + (out[j] + S + sumT[i] - sumT[j]) * (sumF[i] - sumF[j])}。

但是这个方程中有两个变量（out[j] - sumT[j]和sumF[j]），所以无法使用斜率+单调队列优化。

所以改正推为倒推（将前面所说的前缀和改为后缀和）。

新的转移方程：f[i] = max{f[j] + (S + sumT[i] - sumT[j]) * sumF[i]}。

这下使用斜率优化就变得很简单了。

很容易得到斜率不等式：
     f[f] - f[k]
—————————— >= sumF[i]
  sumT[j] - sumT[k]

那么斜率优化就顺理成章了。
Accode：

#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>

const int maxN = 10010;
typedef long long int64;
int64 F[maxN], sT[maxN], sF[maxN], S;
int q[maxN], n, f = 0, r = 1;

inline int getint()
{
    int res = 0; char tmp;
    while (!isdigit(tmp = getchar()));
    do res = (res << 3) + (res << 1) + tmp - '0';
    while (isdigit(tmp = getchar()));
    return res;
}

#define check(j, k, i) \
(F[j] - F[k] <= sF[i] * (sT[j] - sT[k]))

#define cmp(j, k, i) \
((F[j] - F[k]) * (sT[k] - sT[i]) \
<= (F[k] - F[i]) * (sT[j] - sT[k]))

int main()
{
    freopen("batch_scheduling.in", "r", stdin);
    freopen("batch_scheduling.out", "w", stdout);
    n = getint(); S = getint();
    for (int i = 1; i < n + 1; ++i)
    {
        sT[i] = getint(); sF[i] = getint();
    }
    sT[n + 1] = sF[n + 1] = 0;
    q[0] = n + 1;
    for (int i = n; i; --i)
    {
        sT[i] += sT[i + 1]; sF[i] += sF[i + 1];
        while (f < r - 1 && !check(q[f], q[f + 1], i)) ++f;
        F[i] = F[q[f]] + (S + sT[i] - sT[q[f]]) * sF[i];
        while (f < r - 1 && !cmp(q[r - 2], q[r - 1], i)) --r;
        q[r++] = i;
    }
    printf("%I64d\n", F[1]);
    return 0;
}