hdu 1024 Max Sum Plus Plus(m段最大和)

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

思路：

状态dp[i][j]

有前j个数，组成i组的和的最大值。

决策： 第j个数，是在第包含在第i组里面，还是自己独立成组。

方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j

空间复杂度，m未知，n<=1000000，  继续滚动数组。

时间复杂度 n^3. n<=1000000.  显然会超时，继续优化。

max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个

的最大值 用数组保存下来  下次计算的时候可以用，这样时间复杂度为 n^2.

#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int m,n;
int a[1000007];
int dp1[1000007];
int dp2[1000007];
int main(){
    //ios::sync_with_stdio(false);
    while(~scanf("%d%d",&m,&n)){
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int maxn;
        for(int i=1;i<=m;i++){ //枚举子序列
            maxn=-inf;
            for(int j=i;j<=n;j++){ //j = i是因为每个子序列最少1个元素
                dp1[j]=max(dp2[j-1]+a[j],dp1[j-1]+a[j]);
                dp2[j-1]=maxn;
                maxn=max(maxn,dp1[j]);
            }    
        }
        cout<<maxn<<endl;    
    }
}