HDU 1024 Max Sum Plus Plus (dp 滚动数组)

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本文解析了HDU 1024 MaxSumPlusPlus问题，通过动态规划优化解决求m段不相交子段和的最大值问题。采用滚动数组减少空间复杂度，并通过预处理最大值加速计算。

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42781 Accepted Submission(s): 15463

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.

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1024

题目大意：求m段不相交的子段的和

题目分析：容易想到二维dp，dp[i][j]表示分了i段且第i段以数字a[j]结尾的最大值，则转移方程为

dp[i][j] = max(dp[i][j - 1], max(dp[i - 1][k])) + a[j]，空间nm，时间n^2m，显然不可做

容易发现dp[i]的值只和dp[i-1]有关，采用滚动数组优化空间，可同时优化掉一维时间，因为max(dp[i - 1][k])的值可以在i-1阶段求出

#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
int const INF = 0x7fffffff;
int const MAX = 1000005;
int n, m;
ll a[MAX], dp[MAX], pre[MAX], ma;

int main() {
    while (scanf("%d %d", &m, &n) != EOF) {
        for (int i = 1; i <= n; i++) {
            scanf("%I64d", &a[i]);
            dp[i] = pre[i] = 0;
        }
        for (int i = 1; i <= m; i++) {
            ma = -INF;
            for (int j = i; j <= n; j++) {
                dp[j] = max(dp[j - 1], pre[j - 1]) + a[j];
                pre[j - 1] = ma;
                ma = max(ma, dp[j]);
            }
        }
        printf("%I64d\n", ma);
    }
}