HDU 1024:max Sum plus Plus (DP)

http://acm.hdu.edu.cn/showproblem.php?pid=1024Max Sum plus Plus

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 24675 acce pted Submission (s): 8478

problem DescriptionNow I Think you has got an AC in IGNATIUS.L ' s "Max Sum" problem. To is a brave acmer, we always challenge ourselves to more difficult problems. Now you is faced with a more difficult problem.
Given a consecutive number sequence S1, S2, S3, S4 ... Sx... Sn (1≤x≤n≤1,000,000, -32768≤sx≤32767). We define a function sum (i, j) = Si + ... + Sj (1≤i≤j≤n).
Now given a integer m (M > 0), your task is to find m pairs of I and J which make sum (I1, J1) + sum (i2, J2) + SUM ( i3, J3) + ... + sum (IM, JM) maximal (IX≤IY≤JX or IX≤JY≤JX is not allowed).
But i ' m lazy, I don't want to the write a Special-judge module, so you don ' t has to output m pairs of I and J, just OUTPU T the maximal summation of sum (ix, JX) (1≤x≤m) instead. ^_^  InputEach test case would begin with a integers m and n, followed by n integers S1, S2, S3 ... Sn. Process to the end of file. OutputOutput The maximal summation described above in one line.  Sample Input1 3 1 2 6-1 4-2 3-2 3  Sample Output6 8 HintHuge Input, scanf and dynamic programming is recommended. Test Instructions: Divides a string of numbers into M-series, seeking the maximum possible sum.

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <iostream>5 using namespacestd;6 #defineN 10000057 #defineINF-10000000008 intNum[n],dp[n],mmax[n];9 /*Ten status Dp[i][j] One The number of the first J can be divided into the maximum of the I group A Dp[i][j]=max (Dp[i][j-1]+num[j], Max (Dp[i-1][k]) +num[j]),0<k<j; - Dp[i][j-1] is to add num[j] to the front of a group, - Max (Dp[i-1][k]) +num[j] is to re-divide the num[j into another group, the Max (Dp[i-1][k]) is the maximum value that is divided into I-1 groups - open a Mmax array each time can be saved into I-1 groups when using the first j-1 number of times the maximum value - time Complexity O (MN) - */ + intMain () - { +     intn,m; A      while(~SCANF ("%d%d",&m,&N)) { at          for(intI=1; i<=n;i++){ -scanf"%d", num+i); -         } -Memset (DP,0,sizeof(DP)); -memset (Mmax,0,sizeof(Mmax)); -         intans; in          for(intI=1; i<=m;i++){ -ans=INF; to              for(intj=i;j<=n;j++){ +Dp[j]=max (dp[j-1],mmax[j-1])+Num[j]; -mmax[j-1]=ans; theans=Max (Dp[j],ans); *             } $         }Panax Notoginsengcout<<ans<<Endl; -     } the     return 0; +}

2016-06-21

HDU 1024:max Sum plus Plus (DP)