51nod 1052 Max M sub-segment and & POJ 2479 maximum two sub-segments and

extension of the maximum sub-segment and classic problem. title Link: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1052 n integers consisting of a sequence of a[1],a[2],a[3],..., A[n], dividing the n number into disjoint m-sub-segments, and the largest of the M segments. If M >= the number of positive numbers in N, then the sum of all positive numbers is output. For example:-2 11-4 13-5 6-2, divided into 2 segment, 11-4 131 segment, 61 segment, and 26. Input

Line 1th: 2 numbers n and m, separated by a space in the middle. n is the number of integers, and M is divided into how many segments. (2 <= N, M <= 5000) line 2-n+1: N integers ( -10^9 <= a[i] <= 10^9)

Output

Output this most Yamato

Input example

7 2-211-413-56-2

Output example

26

use two arrays, PRE[MAXN] and DP[MAXN]. first the M-cycle, the X-cycle represents the maximum X sub-segments that can be obtained by dividing the entire sequence into X-segments. The Pre[i] Array records the maximum number of sub-segments that can be obtained from the 1th count to the number of I to be divided into X segments. assume that you are currently partitioned into X sub-segments (that is, the outermost loop executes x times), and then you need to perform the first x+1:The Dp[i] transfer is the max value from the following two scenarios. 1. The number of the former i-1 is divided into X sub-paragraphs to obtain the maximum and pre[i-1] plus a separate input[i] as the beginning of the paragraph x+1 sub-paragraph;2. The x+1 sub-paragraph has been separated from the number of i-1, and the tail of the X+1 sub-paragraph needs to be added input[i], i.e. dp[i-1]+input[i] AC Code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < Limits.h>using namespace Std;typedef long long LL; #define MAXN 5010const ll min_inf =-(1 << 30);    LL DP[MAXN], INPUT[MAXN], Pre[maxn];int N, m;int Main () {scanf ("%d%d", &n, &m);    Dp[0] = Min_inf;    Pre[0] = 0;        for (int i = 1; I <= n; i++) {scanf ("%i64d", &input[i]);        Pre[i] = 0;    Dp[i] = Min_inf;    } LL ans = min_inf;        while (m--) {ans = min_inf;            for (int i = 1; I <= n; i++) {if (i = = 1) dp[i] = Pre[i-1] + input[i];                else {if (dp[i-1] > Pre[i-1]) {dp[i] = Dp[i-1] + input[i];//The number of I and its previous number within the same subsection            } else {Dp[i] = Pre[i-1] + input[i];//A new sub-segment from I}} Dp[i] = max (dp[i-1], pre[i-1]) + input[i];            Or the direct value can be pre[i-1] = ans; ans = max (ans, dp[i));    } Pre[n] = ans;    } printf ("%i64d\n", Pre[n]); return 0;}

　　

Similarly, m=2 can also solve the problem of poj2479. http://poj.org/problem?id=2479

51nod 1052 Max M sub-segment and & POJ 2479 maximum two sub-segments and