Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20887 Accepted Submission(s): 6960

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.

题目大意：在一个元素个数为m的序列中取出互不交叉的n段，找出最大值

dp[i - 1][j - 1]表示前j - 1个数取i - 1段得到的最大和，第j个数a[j]有两种情况：

1.a[j]单独作为一段,为第i段；

2.a[j]是紧跟第j - 1个数a[j - 1]之后,仍是第i - 1段的一部分。

由于数据太大所以将二维数组转化为一维数组

Max[j]表示上一个状态最终得到的最大值

dp[j] = max(dp[j - 1] + a[j], Max[j - 1] + a[j]);

dp[j - 1] + a[j] 表示情况2 （a[j]是紧跟第j - 1个数a[j - 1]之后,仍是第i - 1段的一部分）

Max[j - 1] + a[j] 表示情况1 ( a[j]单独作为一段,为第i段)

详解亲请参考：

http://blog.csdn.net/a342374071/article/details/6701544

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 1000010

using namespace std;

int dp[N], Max[N], a[N];

int main()
{
    int n, m, sum;
    while(~scanf("%d%d", &n, &m))
    {
        int i, j;
        memset(dp, 0, sizeof(dp));
        memset(Max, 0, sizeof(Max));
        for(i = 1 ; i <= m ; i++)
            scanf("%d", &a[i]);
        for(i = 1 ; i <= n ; i++)//i表示选取的段数
        {
            sum = -INF;
            for(j = i ; j <= m ; j++)//j表示元素的个数
            {
                dp[j] = max(dp[j - 1] + a[j], Max[j - 1] + a[j]);//此处见上面解释
                Max[j - 1] = sum;//Max[j - 1]记录j - 1这一状态得到的最大值
                sum = max(sum, dp[j]);
            }
            Max[j - 1] = sum;//状态更新
        }
        printf("%d
", sum);
    }
    return 0;
}