2022.11.23Max Sum Plus Plus HDU - 1024

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 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample

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

6 8

Hint

Huge input, scanf and dynamic programming is recommended.

原题链接：传送门 

题意：长度为n的序列，m个子段和的最大值

思路：我们构造dp[i][j]状态表示为前j个数分成i组的最大值

状态转移方程：

我们看到数据范围会发现二维dp不太可行，所以我们需要对二维dp进行压缩，可以另开一个数组来代表dp[i-1][k]这样我们就能放心地去压缩一维。

ps:在循环中f[j-1]代表前j-1个数分成i-1组的最大值

#include 
using namespace std;
typedef long long LL;
const int inf = 1 << 30;
int main(){
  int n, m;
  while(~scanf("%d%d", &m, &n)){
    vector a(n + 1), f(n + 1, 0), dp(n + 1, 0);
    for(int i = 1; i <= n; i++) cin >> a[i];
    dp[0] = -inf;LL ans;
    for(int i = 1; i <= m; i++){
      ans = -inf;
      for(int j = 1; j <= n; j ++){
        dp[j] = max(dp[j - 1] +  a[j], f[j - 1] + a[j]);
        f[j - 1] = ans;
        ans = max(ans, dp[j]);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}