HDOJ1024--Max Sum Plus Plus(动态规划)UnSolved

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 S ₁, S ₂, S ₃, S ₄ ... 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(i ₁, j ₁) + sum(i ₂, j ₂) + sum(i ₃, j ₃) + ... + 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. ^_^ 

InputEach test case will begin with two integers m and n, followed by n integers S ₁, S₂, S ₃ ... S _n. 
Process to the end of file. 
OutputOutput 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.

       

若不做任何优化，并不考虑数据大小，仅考虑样例

#include<iostream>
#include<algorithm>
using namespace std;
int num[1000];
int dp[10][100];
int main(){
    int n,m;
    while(cin>>n>>m){
        for(int i=1;i<=m;i++){
            cin>>num[i];
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                int m=-1;
                for(int w=1;w<j;w++){
                    m=max(dp[i-1][w],m);
                }
                dp[i][j]=max(m,dp[i][j-1])+num[j];
            }
        }
        int ans=-1;
        for(int i=1;i<=m;i++){
            ans=max(ans,dp[n][i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int dp[1000005];
int Max[1000005];
int num[1000005];
int main(){
    int m,n;
    while(cin>>m>>n){
        int k;
        for(int i=1;i<=n;i++){
            cin>>num[i];
        }
        memset(dp,0,sizeof(dp));
        memset(Max,0,sizeof(Max));
        int mmax;
        
        for(int i=1;i<=m;i++){
            mmax=-INT_MAX;
            for(int j=i;j<=n;j++){
                dp[j]=max(dp[j-1],Max[j-1])+num[j];
                Max[j-1]=mmax;
                mmax=max(mmax,dp[j]);
            }
        }
        cout<<mmax<<endl;
    }
    return 0;
}