HDU1024 Max Sum Plus Plus （优化线性dp）

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. ^_^

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

6
8


        
 

Huge input, scanf and dynamic programming is recommended.

#include <bits/stdc++.h>
#define INF 1<<30
using namespace std;
int a[1000005];
int dp[1000005]={0};
int lmax[1000005]={0};
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        int i,j;
        memset(dp,0,sizeof(dp));
        memset(lmax,0,sizeof(lmax));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans=0;
        for(i=1;i<=m;i++)
        {
            ans=-INF;
            for(j=i;j<=n;j++)
            {
                dp[j]=max(dp[j-1]+a[j],lmax[j-1]+a[j]);
                lmax[j-1]=ans;
                ans=max(ans,dp[j]);
            }
        }
        cout<<ans<<endl;
    }    
    return 0;
}