HDU 1024 Max Sum Plus Plus(二维数组转化为一维数组)

Problem Description：

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 S₁, S₂, S₃, S₄ ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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. ^_^

Input：

Each test case will begin with two integers m and n, followed by n integers S₁, S₂, S₃ ... S_n.
Process to the end of file.

Output：

Output 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：

Hint

Huge input, scanf and dynamic programming is recommended.

题意：有一个长度为n的序列，现在要被分成不交叉的m段序列（不用分完所有的元素），问怎样分才能使这m段序列的和最大，这道题不求这m段是什么，只用输出这个最大的值。

假设我们让数组a[j]表示输入的数组，DP[i][j]表示前j个数被分成i组的最大和（不用分完所有前j个元素），那么我们必须明白：第j个数可能被独立分成第i组，也可能加在第j-1个数的后面，成为第i组的元素（此时第i组元素只有第j个数），所以状态转移方程变为dp[j] = max(dp[j-1]+a[j], Max[j-1]+a[j]) （因为如果开二维数组内存过大，所以我们可以采用双重for循环形式进行计算最大值，此时只需要两个一维数组就可以了）。

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;

const int N=1000010;
const int INF=0x3f3f3f3f;

int a[N], dp[N], Max[N]; ///dp[j]存放前j个数分组后的最大值，Max[j-1]存放上一组的最大值

int main ()
{
    int m, n, i, j, M;

    while (scanf("%d%d", &m, &n) != EOF)
    {
        for (i = 1; i <= n; i++)
            scanf("%d", &a[i]);

        memset(dp, 0, sizeof(dp));
        memset(Max, 0, sizeof(Max));

        for (i = 1; i <= m; i++) ///i代表组数，j代表元素个数
        {
            M = -INF; ///每次多分一组都需要统计最大值

            for (j = i; j <= n; j++) ///j从i开始是因为组数肯定要比元素个数少
            {
                dp[j] = max(dp[j-1]+a[j], Max[j-1]+a[j]); ///dp[j-1]+a[j]代表a[j]连在a[j-1]后面成为第i组，但是该组只有a[j]一个元素，Max[j-1]+a[j]代表a[j]独立成组
                Max[j-1] = M; ///先保存最大值，后计算这一组的最大值，则下次使用Max数组时就变成了上一组的最大值了
                M = max(M, dp[j]);
            }
        }

        printf("%d
", M);
    }

    return 0;
}