HDU 1024 Max Sum Plus Plus
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13897 Accepted Submission(s):
4569
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. ^_^
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.
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.
Author
JGShining(极光炫影)
思路:DP,使用二维数组会超空间,所以我选择滚动数组
代码:
#include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; int n,m; int map[1000010]; long long dp[2][100000]; long long maxn[100000]; int the_last_sum; int max(int x,int y) { return x > y ? x : y; } int main() { while(~scanf("%d%d",&m,&n)) { memset(map,0,sizeof(map)); for(int i = 1;i <= m;i ++) maxn[i] = -999999999; for(int i = 0;i <= 2;i ++) for(int j = 1;j <= m;j ++) dp[i][j] = -999999999; for(int i = 1;i <= n;i ++) scanf("%d",&map[i]); dp[1][1] = map[1]; for(int i = 1;i <= n;i ++) for(int j = m;j >= 1;j --) { dp[i % 2][j] = max(dp[(i - 1) % 2][j] + map[i],maxn[j - 1] + map[i]); if(dp[i % 2][j] > maxn[j]) maxn[j] = dp[i % 2][j]; } printf("%lld\n",maxn[m]); } return 0; }