poj1180 Batch Scheduling
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 3590 | Accepted: 1654 |
Description
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
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
Sample Input
5
1
1 3
3 2
4 3
2 3
1 4
Sample Output
153
这题顺着推很难推,因为对于dp[i],查找的dp[k]和前k个部件运完后的时间不知道,而这两者都会影响dp[i],所以考虑倒着推。可以发现每个任务对对最后代价的贡献实际上等于它及它以后的f之和乘以它的时间t,即后面的任务都要为它等上t的时间,会多花f*t的代价。用dp[i]表示i到n部件运送完后所花的最小价值,用st[i]表示i到n的所有时间和,sf[i]表示i到n的所有价值和,那么dp[i]=min(dp[k]+(st[i]-st[k]+s)*f[i]).
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 10050
int dp[maxn];
int t[maxn],v[maxn];
int q[1111111];
int main()
{
int n,m,i,j,s;
while(scanf("%d",&n)!=EOF)
{
scanf("%d",&s);
for(i=0;i<=n;i++)dp[i]=inf;
dp[0]=0;
for(i=1;i<=n;i++){
scanf("%d%d",&t[i],&v[i]);
}
reverse(t+1,t+1+n);
reverse(v+1,v+1+n);
t[0]=v[0]=0;
for(i=1;i<=n;i++){
t[i]=t[i-1]+t[i];
v[i]=v[i-1]+v[i];
}
int front,rear;
front=rear=1;
q[rear]=0;
for(i=1;i<=n;i++){
while(front<rear && (dp[q[front+1] ]-dp[q[front] ]<=v[i]*(t[q[front+1] ]-t[q[front] ]) ) ){
front++;
}
int k=q[front];
dp[i]=dp[k]+(s+t[i]-t[k])*v[i];
while(front<rear && (dp[q[rear] ]-dp[q[rear-1] ])*(t[i]-t[q[rear] ])>=(dp[i]-dp[q[rear] ] )*(t[q[rear] ]-t[q[rear-1] ]) ){
rear--;
}
rear++;
q[rear]=i;
}
printf("%d\n",dp[n]);
}
return 0;
}
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