Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24152 Accepted Submission(s): 8265
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
经典的动态规划+滚动数组题
#include<iostream>
#define max(a,b) (a>b?a:b)
#define MAX 1000000
#define INF 0x7fffffff
using namespace std;
int main(void)
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
int n, m;
while (cin >> n >> m)
{
int d[MAX+1];
int dp[MAX + 1] = { 0 }; //前J个数，组成I组最大
int mm[MAX+1] = { 0 };
int mmax;
for (int i = 1; i<=m; i++)
{
cin >> d[i];
}
for (int i = 0; i < n; i++)
{
mmax = -INF;
for (int j = i+1; j <= m; j++)
{
dp[j] = max(dp[j - 1], mm[j - 1]) + d[j];
mm[j - 1] = mmax;
mmax = max(dp[j], mmax);
}
}
cout << mmax << endl;
}
return 0;
}