HDU1024-Max Sum Plus Plus【动态规划】

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 ≤ iy ≤ 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, S2, S 3 ... S n. 
Process to the end of file. 

Output

Output the maximal summation described above in one line. 

思路：我们用一个数组来记录每层的最大值，这样就省去了一层循环，最后结果取最优值。

#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
int a[maxn], dp[maxn], Max[maxn];
int main()
{
    int n, m;
    while(~scanf("%d%d", &m, &n))
    {
        for(int i = 1; i <= n; ++i)
            scanf("%d", &a[i]);
        memset(dp, 0, sizeof(dp));
        memset(Max, 0, sizeof(Max));
        int ans;
        for(int i = 1; i <= m; ++i)
        {
            ans = -inf;
            for(int j = i; j <= n; ++j)
            {
                dp[j] = max(dp[j - 1], Max[j - 1]) + a[j];
                Max[j - 1] = ans;
                ans = max(ans, dp[j]);
            }
        }
        cout << ans << endl;
    }
    return 0;
}