Vijos P1243 Production products (monotone queue +DP)

P1243 Production ProductsDescribe

After a period of operation, Dd_engi's Oi shop is not satisfied with the purchase of products from other suppliers to put on shelves, but to begin to produce their own products! The production of the product requires a M-step, each of which can be done on any one of the N machines, but the production steps must be executed in strict order. Because of the different performance of the N machines, the time required to complete each step is different. Machine I completed the time of the J step is t[i,j]. It also takes a certain amount of time to move semi-finished products from one machine to another machine. At the same time, in order to ensure the safety and quality of the product, each machine can only be completed in a continuous L-step product. That is, if a machine has completed the L-step of the product in succession, the next step must be done with a different machine. Now, the first product of Dd_engi's Oi store is about to start production, so how long will it take?
One day azuki.7 to the movement said: This topic is too simple, we change the scope of the topic to change
This is a difficult question for rookie, and he wants you to help him solve it.

Format input Format

The first line has four integers m, N, K, L
The following n rows have m integers per line. The J integer of Line i+1 is t[j,i].

Output format

The output has only one row, indicating the shortest time required.

Example 1 sample input 1[copy]

3 2 0 22 2 31 3 1

Sample output 1[Copy]

4

Limit

1s

Tips

For 50% of data, n<=5,l<=4,m<=10000
For 100% of data, n<=5, l<=50000,m<=100000

Source

First "Oi Shop Cup" dd_engi original topic

Ideas

The monotone queue optimizes DP.

F[I][J] indicates that the first step is taken into account, and the minimum time at which I step is completed by J. With Transfer type:

F[i][j]=min{f[k][p]-sum[k][j]}+sum[i][j] , I-l<=k<=i-1

The parts in parentheses can be maintained with n monotone queues. A[][] is an auxiliary array,

A[i][j]=min{f[i][k]}-sum[i][j]+k,1<=k<=m

Represents F[I][J] The minimum value that can be taken in parentheses, and the minimum value of the monotonic queue J maintenance interval [i-l,i-1] a[][j].

Note the monotone queue notation.

PS: This paper is good for beginners: HTTP://PAN.BAIDU.COM/S/1GEXNMWJ

Code

1#include <cstdio>2#include <cstring>3#include <iostream>4 using namespacestd;5 6 Const intMAXN =200000+Ten;7 Const intINF = 1e9+1e9;8 9 intl[6],r[6],q[maxn][6];Ten intf[maxn][6], a[maxn][6], b[maxn][6]; One intN,m,k,lim; A  - voidPushintIintj) { -      while(L[j]<r[j] && a[i][j]<=a[q[r[j]-1][J]][J]) r[j]--; theq[r[j]++][j]=i; - } - intPopintIintj) { -      while(L[j]<r[j] && Q[l[j]][j]<i-lim) l[j]++; +     returnA[q[l[j]][j]][j]; - } + intReadint&x) { A     CharC=getchar (); while(!isdigit (c)) c=GetChar (); atx=0; while(IsDigit (c)) x=x*Ten+c-'0', c=GetChar (); - } - intMain () { - Read ( N), read (m), read (K), read (LIM); -      for(intI=1; i<=m;i++) -          for(intj=1; j<=n;j++) { in read (b[j][i]); -b[j][i]+=b[j-1][i]; to         } +      for(intj=1; j<=m;j++) Push (0, j); -      for(intI=1; i<=n;i++) the     { *          for(intj=1; j<=m;j++) $F[i][j]=pop (I,J) +B[i][j];Panax Notoginseng          for(intj=1; j<=m;j++) {//add a[i][to the monotone queue] -a[i][j]=INF; the              for(intk=1; k<=m;k++)if(k!=j) +a[i][j]=min (a[i][j],f[i][k]); Aa[i][j]=a[i][j]-b[i][j]+K; the push (I,J); +         } -     } $     intans=f[n][1]; $      for(intI=1; i<=m;i++) ans=min (ans,f[n][i]); -printf"%d\n", ans); -     return 0; the}

Vijos P1243 Production products (monotone queue +DP)