hdu 1024 Max Sum Plus Plus (dp+二维变一维)

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13722    Accepted Submission(s): 4522

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 32 6 -1 4 -2 3 -2 3

 

Sample Output

68
Hint
Huge input, scanf and dynamic programming is recommended.
 

 

思路：

dp。

状态：dp[i][j]-最大和    i-已经组成多少组  j-已经用掉多少数。

方程：dp[i][j]=max(dp[i-1][j]+s[i],dp[k][j-1]+s[i]);   (j>=i)  (0<k<i)

时间复杂度为O(n*m^2),dp[k][j-1]可以递推时用一个值维护，降到O(n*m).

由于n比较大，空间吃不消，所以要优化空间。

考虑到第一维每次都是从i-1到i，故可以将第一维去掉。

方程变为 dp[j]=max(dp[j-1]+s[j],ma[j-1]+s[j]);

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define maxn 1000005#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;ll n,m,ans;ll s[maxn],ma[maxn];ll dp[maxn];void solve(){    int i,j;    ll t;    ma[0]=dp[0]=0;    for(i=1;i<=m;i++)    {        t=-INF;        for(j=i;j<=n;j++)        {            dp[j]=max(dp[j-1]+s[j],ma[j-1]+s[j]);            ma[j-1]=t;            t=max(t,dp[j]);        }        ma[n]=t;    }}int main(){    int i,j;    while(~scanf("%I64d%I64d",&m,&n))    {        for(i=1;i<=n;i++)        {            scanf("%I64d",&s[i]);            ma[i]=0;        }        solve();        printf("%I64d\n",ma[n]);    }    return 0;}/*1 31 2 32 6-1 4 -2 3 -2 32 6-1 -2 -3 -4 -5 -6*/