HDU 1024 Max Sum Plus Plus【DP+优化时间复杂度】

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.

 

思路

思路很清晰,但是超时掉,寻找优化算法. 题意：是找K段序列和使他的和最大。 状态转移方程为： f[i][j]表示连续j个数分成i段最大            f[i-1][j-1]+a[i]    当i==j时 f[i][j]=             max(f[i-1][k])   i-1=<k<=j-1 但是这样算法的时间复杂度达到O（n^3）会超时，因此再算出来 f[i][j]时就把最大的f[i][k]算出来，时间复杂度为O（n^2）

源码

#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; #define INF 999999999 int a[1000023]; __int64 f[2][1000023]; int main() {     int i, j, n, m, k;     while(scanf("%d%d", &m, &n)!=EOF)     {         //memset(a, 0, sizeof(a));         //memset(f, 0, sizeof(f));  //这里会导致超时         for(i=1; i<=n; i++)         {             scanf("%d", &a[i]);             f[1][i]=f[0][i]=0;         }         int t=1;         for(i=1; i<=m; i++)         {             f[t][i]=f[1-t][i-1]+a[i];             f[1-t][i]=f[1-t][i-1]>f[1-t][i]?f[1-t][i-1]:f[1-t][i];             for(j=i+1; j<=n; j++)            {                 f[t][j]=(f[t][j-1]>f[1-t][j-1]?f[t][j-1]:f[1-t][j-1])+a[j];                 f[1-t][j]=f[1-t][j]>f[1-t][j-1]?f[1-t][j]:f[1-t][j-1];             }             t=1-t;         }         t=1-t;         __int64 maxnum=-INF;         for(i=m; i<=n; i++)             maxnum=max(maxnum, f[t][i]);         printf("%d\n", maxnum);     } }