【HDU 1024 Max Sum Plus Plus】+ dp

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本文介绍了一个名为 MaxSumPlusPlus 的问题，该问题要求在给定序列中找到 m 对下标，使得这些下标的子序列之和最大。通过动态规划的方法解决这一挑战，关键在于维护两个数组 dp 和 ans 来分别记录当前最优解和上一步的最优解。

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30112 Accepted Submission(s): 10593

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.

Author
JGShining（极光炫影）

思路： dp，记录上次的最优解和本次的最优解

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e6 + 10;
const int INF = 0x3f3f3f3f;
typedef long long LL;
LL dp[MAX],ans[MAX],a[MAX],sum;
int main()
{
    int m,n;
    while(~scanf("%d %d",&m,&n)){
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        for(int i = 1; i <= n; i++)
            scanf("%lld",&a[i]);
        for(int i = 1; i <= m; i++){
            sum = -INF;
            for(int j = i; j <= n; j++) // ans 数组只有在下次循环才会用到
                dp[j] = max(dp[j - 1],ans[j - 1]) + a[j],ans[j - 1] = sum,sum = max(sum,dp[j]);
        }
        printf("%lld\n",sum);
    }
    return 0;
}