HDU1024--Max Sum Plus Plus

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.

 

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define LL long long int#define inf 0x7fffffff#define maxn 1000080dp[i][j] = dp[i-1][j-1] + a[i],dp[i-1][j]...也就是说。看这个数大于0，还是小于0先。int dp[maxn],A[maxn],ffuck[maxn];inline int max(int a,int b){return a>b?a:b;}int main(){int n,m;while(scanf("%d%d",&m,&n)==2){for(int i = 1;i <= n;i++)scanf("%d",&A[i]);for(int i = 0;i <= n;i++){dp[i] = 0;ffuck[i] = 0;}int fuck;for(int i = 1;i <= m;i++){fuck = -inf;for(int j = i;j <= n;j++){dp[j] = max(dp[j-1] + A[j],ffuck[j-1] + A[j]);ffuck[j-1] = fuck;fuck = max(dp[j],fuck);}}printf("%d\n",fuck);}return 0;}