Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39886 Accepted Submission(s): 14338

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...Sx,...Sn(1≤x≤n≤1,000,000,−32768≤Sx≤32767)S_1, S_2, S_3, S_4 ... S_x, ... S_n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S_x ≤ 32767)S1​,S2​,S3​,S4​...Sx​,...Sn​(1≤x≤n≤1,000,000,−32768≤Sx​≤32767). We define a function sum(i,j)=Si+...+Sj(1≤i≤j≤n)sum(i, j) = S_i + ... + S_j (1 ≤ i ≤ j ≤ n)sum(i,j)=Si​+...+Sj​(1≤i≤j≤n).

Now given an integer m(m>0)m (m > 0)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(im,jm)sum(i_1, j_1) + sum(i_2, j_2) + sum(i_3, j_3) + ... + sum(i_m, j_m)sum(i1​,j1​)+sum(i2​,j2​)+sum(i3​,j3​)+...+sum(im​,jm​) maximal (ix≤iy≤jxi_x ≤ i_y ≤ j_xix​≤iy​≤jx​ or ix≤jy≤jxi_x≤ j_y ≤ j_xix​≤jy​≤jx​ 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 iii and jjj, just output the maximal summation of sum(ix,jx)(1≤x≤m)sum(i_x, j_x)(1 ≤ x ≤ m)sum(ix​,jx​)(1≤x≤m) instead.

Input

Each test case will begin with two integers mmm and nnn, followed by nnn integers S1,S2,S3...SnS_1, S_2, S_3 ... S_nS1​,S2​,S3​...Sn​.
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

题意

将一个长度为nnn的数组分成不相交的mmm段，求这mmm段的和的最大值

思路

状态：dp[i][j]dp[i][j]dp[i][j]表示在前jjj个数中取出iii段的最大和

状态转移方程：dp[i][j]=max(dp[i−1][k],dp[i][j−1])+num[j]  (i−1≤k≤j−1)dp[i][j]=max(dp[i-1][k],dp[i][j-1])+num[j]\ \ (i-1\leq k \leq j-1)dp[i][j]=max(dp[i−1][k],dp[i][j−1])+num[j]  (i−1≤k≤j−1)

由于mmm范围未知，n≤106n\leq 10^6n≤106，所以二维的dp方程无论是在时间上还是在空间上都是不允许的。

那么我们就需要对这个方程进行优化：

不难发现当前状态只与两个状态有关：

1. 第jjj个数和前j−1j-1j−1个数在一段里
2. 第jjj个数和前j−1j-1j−1个数不在一段里。

根据这一点，我们把状态降成一维的数组，dp[j]dp[j]dp[j]表示前jjj个数分iii段时的最大和，然后用sum[j−1]sum[j-1]sum[j−1]来表示状态一的前j−1j-1j−1个数在前i−1i-1i−1段的最大和，dp[j−1]dp[j-1]dp[j−1]表示状态二的前j−1j-1j−1个数在前iii段的最大和。

当前状态的转移方程为：dp[j]=max(dp[j−1],sum[j−1])+num[j]dp[j]=max(dp[j-1],sum[j-1])+num[j]dp[j]=max(dp[j−1],sum[j−1])+num[j]，持续更新dp与sum数组的值

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#include <time.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
#define bug cout<<"-------------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int a[maxn];
int dp[maxn];
int sum[maxn];
int main(int argc, char const *argv[])
{ios::sync_with_stdio(false);cin.tie(0);#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);double _begin_time = clock();#endifint k,n;while(cin>>k>>n){int res;for(int i=1;i<=n;i++)cin>>a[i];ms(dp,0);ms(sum,0);for(int i=1;i<=k;i++){res=-INF;for(int j=i;j<=n;j++){dp[j]=max(sum[j-1],dp[j-1])+a[j];sum[j-1]=res;res=max(res,dp[j]);}}cout<<res<<endl;}#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %lf ms.", _end_time - _begin_time);#endifreturn 0;
}