Max Sum Plus Plus HDU - 1024

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are 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 an integer m (m > 0), your task is to find m pairs of i and j which make sum(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + sum(i _m, j _m) maximal (i _x ≤ i_y ≤ j _x or i _x ≤ j _y ≤ j _x is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i _x, j _x)(1 ≤ x ≤ m) instead. ^_^

InputEach test case will begin with two integers m and n, followed by n integers S ₁, S₂, S ₃ ... S _n.
Process to the end of file.
OutputOutput the maximal summation described above in one line.
Sample Input

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8

Hint

Huge input, scanf and dynamic programming is recommended.

本题的大致意思为给定一个数组，求其分成m个不相交子段和最大值的问题。

用一个数组去寻找当前段的最大值，用另一个数组保存寻找过的最大值。

// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#define INF 0x3f3f3f3f
#define mod 2016
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1000000+5;
int n, m, T, len, cnt, num, Max;
int dp[maxn], mmax[maxn];
int a[maxn];

void input() {
    while( scanf("%d%d", &m, &n)!=EOF ) {
        for(int i=1; i<=n; i++) {
            scanf("%d", &a[i]);
            dp[i] = 0;
            mmax[i] = 0;
        }
        dp[0] = 0;
        mmax[0] = 0;
        for(int i=1; i<=m; i++) {
            Max = -INF;
            for(int j=i; j<=n; j++) {
                dp[j] = max(dp[j-1]+a[j], mmax[j-1]+a[j]);
                mmax[j-1] = Max;
                Max = max(Max, dp[j]);
            }
        }
        printf("%d\n", Max);
    }
}

int main() {
    input();
    return 0;
}