HDU Max Sum plus Plus (recursive)

Max Sum plus Plus

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 18653 Accepted Submission (s): 6129

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, S 2, S 3, S 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, S 2, S 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 HintHuge input, scanf and dynamic programming is recommended.

Dp[i][j][0] ... Indicates that the number of first I is divided into J groups, and the maximum score of the number of No. i is not selected

DP[I][J][1] ... Indicates the number of first I is divided into J groups, the maximum score of the number of selected I

Because state i is only i-1 with the state, you can use a scrolling array to reduce space

Take the most to write yourself. Otherwise the card constant will time out

#include <iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespacestd;Const intN =100010;Const intINF = 1e9+7;intdp[2][n][2], N, M, x[n]; inlineintMAX (intAintb) {if(A > B)returnA; Else returnb;}intMain () {//freopen ("In.txt", "R", stdin);     while(~SCANF ("%d%d",&m,&N)) { for(inti =1; I <= N; ++i) {scanf ("%d",&X[i]); }        intv =0 ; dp[v][0][0] =0 ; dp[v][1][1] = x[1] ;  for(inti =1; I < n; ++i) { for(intj =0; J <= i +1&& J <= M; J + +) {Dp[v^1][j][0] = dp[v^1][j][1] = -inf; }             for(intj = min (M, i); J >=0; --j) {if(J! =i) {dp[v^1][j+1][1] = MAX (dp[v][j][0] + x[i+1], dp[v^1][j+1][1] ); Dp[v^1][j][0] = MAX (dp[v][j][0], dp[v^1][j][0]); }                if(J! =0) {Dp[v^1][j][1] = MAX (dp[v^1][j][1], dp[v][j][1] + x[i+1] ) ; Dp[v^1][j+1][1] = MAX (dp[v^1][j+1][1], dp[v][j][1] + x[i+1] ) ; Dp[v^1][j][0] = MAX (dp[v^1][j][0], dp[v][j][1] ) ; }} v^=1 ; }        intAns =-inf; if(M < n) ans = MAX (ans, dp[v][m][0] ); if(M >0) ans = MAX (ans, dp[v][m][1] ); printf ("%d\n", ans); }    return 0 ;}

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HDU Max Sum plus Plus (recursive)