HDU 1024 Max Sum Plus Plus（基础dp)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34541    Accepted Submission(s): 12341

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）

 

Recommend

We have carefully selected several similar problems for you:  1074 1025 1081 1080 1160 

 

原题地址

 

题意：给我们一个长度为N的数组让我们把数组分成M个不想交的字串 使得M个字串的和最大

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 1100000
#define INF 0x3f3f3f3f
int dp[MAXN];
int maxn[MAXN];
int a[MAXN];
int main()
{
    int n,m;
    std::ios::sync_with_stdio(false);
    while(cin>>m>>n){
        for(int i=1;i<=n;i++){
                cin>>a[i];
                maxn[i]=0;
                dp[i]=0;
        }
        dp[0]=0;
        maxn[0]=0;
        int maxx;
        for(int i=1;i<=m;i++){
                 maxx=-INF;
                for(int j=i;j<=n;j++){
                        dp[j]=max(dp[j-1]+a[j],maxn[j-1]+a[j]);
                        maxn[j-1]=maxx;
                        maxx=max(maxx,dp[j]);
                }
        }
        cout<<maxx<<endl;
    }
    return 0;
}