ACM Dynamic programming Max M sub-segments and problems (in class)

Max Sum plus Plus

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 23976 Accepted Submission (s): 8199

Problem Descriptionnow I Think you have got a 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 S 1+ t 2+ t 3+ t 4... S x, ... S N(1≤x≤n≤1,000,000, -32768≤s x≤32767). We define a function sum (i, j) = S I+ ... + S J(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 (i 1J 1) + SUM (i 2J 2) + SUM (i 3J 3) + ... + sum (i mJ m) Maximal (i x≤i y≤j xOr I x≤j y≤j xis 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 output th e maximal summation of sum (i xJ x) (1≤x≤m) instead. ^_^

Inputeach test case would begin with a integers m and n, followed by n integers S 1+ t 2+ t 3... S N.
Process to the end of file.

Outputoutput the maximal summation described above in one line.

Sample INPUT1 3 1 2 32 6-1 4-2 3-2 3

Sample Output68  

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;Const intMAX = 1e6 +Ten;Const intMIN =-1e8;intCurr[max], Pre[max], A[max];intmax_sum, N, M;intMain () { while(Cin >> M >>N) { for(inti =1; I <= N; i++) Cin>>A[i]; memset (Curr,0,sizeof(Curr)); memset (PRE,0,sizeof(pre)); intj =0;  for(inti =1; I <= m; i++) {max_sum=MIN;  for(j = i; J <= N; j + +) {Curr[j]= Max (Curr[j-1], Pre[j-1]) +A[j];/*Pre[j-1] Represents the maximum value of i...j-1 in the previous state, max_sum the maximum value of the I...J represented after the update, so it cannot be written back*/pre[j-1] =max_sum; Max_sum=Max (Max_sum, Curr[j]); }            //Pre[j-1] The maximum value of the previous state is always maintained, which is importantPre[j-1] =max_sum; } cout<< Max_sum <<Endl; }    return 0;}

ACM Dynamic programming Max M sub-segments and problems (in class)