Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37418 Accepted Submission(s): 13363
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 ... 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 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.
C/C++：
1 #include <bits/stdc++.h>
2 #define INF 0x3f3f3f3f
3 using namespace std;
4
5 const int MAX = 1e6 + 10;
6
7 int m, n, pre[MAX], dp[MAX], num[MAX], ans, j;
8
9 int main()
10 {
11 while (~scanf("%d%d", &m, &n))
12 {
13 memset(dp, 0, sizeof(dp));
14 memset(pre, 0, sizeof(pre));
15
16 for (int i = 1; i <= n; ++ i) scanf("%d", &num[i]);
17 for (int i = 1; i <= m; ++ i)
18 {
19 ans = -INF;
20 for (j = i; j <= n; ++ j)
21 {
22 dp[j] = max(dp[j - 1], pre[j - 1]) + num[j];
23 pre[j - 1] = ans;
24 ans = max(dp[j], ans);
25 }
26 // pre[j - 1] = ans;
27 }
28
29 printf("%d\n", ans);
30 }
31 return 0;
32 }