Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17164 Accepted Submission(s): 5651
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 S 1, S 2, S 3, S 4 ... 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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + 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 S 1, S 2, S 3 ... 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. /*** dp[i][j]表示以第i个数字结尾且选定并分成j份能得到的最大值。转移方程为** dp[i][j] = max(dp[i-1][j], max(dp[1...i-1][j-1])) + arr[i];** 假设开二维数组的话内存会超，所以得用滚动数组省空间。preMax[j]保存** 上一轮得到的dp[1...i][j]中的最大值，ans每次读取当前dp数组最大值** 用以更新preMax数组，最后一轮循环后ans保存的就是答案。*/#include#include #define maxn 1000010#define inf 0x7fffffffint dp[maxn], preMax[maxn], arr[maxn];int max(int a, int b) { return a > b ? a : b;}int main() { int n, m, i, j, ans; while(scanf("%d%d", &n, &m) == 2) { for(i = 1; i <= m; ++i) { scanf("%d", &arr[i]); preMax[i] = dp[i] = 0; } preMax[0] = dp[0] = 0; for(j = 1; j <= n; ++j) { // 分成j份 ans = -inf; for(i = j; i <= m; ++i) { dp[i] = max(dp[i-1], preMax[i-1]) + arr[i]; preMax[i-1] = ans; ans = max(ans, dp[i]); } } printf("%d\n", ans); } return 0;}