poj1180 Batch Scheduling

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3590		Accepted: 1654

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 + (T _x + T _x+1 + ... + T _x+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

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;
}