HDU1024（最大M子段和）

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21540    Accepted Submission(s): 7215

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

6

8

 

Hint

Huge input, scanf and dynamic programming is recommended.

 

Author

JGShining（极光炫影）

 

dp[i][j]表示前j个数组成的序列最大i子段和，则状态转移方程为

dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]} i-1=<t<j-1

优化1：本题数据范围太大， 所以可以用滚动数组优化

优化2：用last数组记录上一层循环得到的最大值，具体见代码

/*
ID: LinKArftc
PROG: 1024.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <climits>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 1000010;

int num[maxn], dp[maxn], last[maxn];

int main() {
    int n, m;
    while (~scanf("%d %d", &m, &n)) {
        memset(dp, 0, sizeof(dp));
        memset(last, 0, sizeof(last));
        for (int i = 1; i <= n; i ++) scanf("%d", &num[i]);
        int ma;
        for (int i = 1; i <= m; i ++) {//分成i段
            ma = INT_MIN;
            for (int j = i; j <= n; j ++) {//前j个数
                dp[j] = max(dp[j-1], last[j-1]) + num[j];//dp[j]表示前j个划分成i段最大值，last[j-1]表示前j-1个划分成i-1段最大值
                last[j-1] = ma;
                ma = max(ma, dp[j]);
            }
            last[n] = ma;
        }
        printf("%d
", ma);
    }

    return 0;
}