hdu 1024 Max Sum Plus Plus (dp)

Max Sum Plus Plus

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 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 ≤ i _y≤ 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, S ₂, S ₃ ... S _n.
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.

 

 

题意：在长度为n的数列中选取k个不重叠的区间，使这k区间和之和最大。输出这个最大值。

 

思路：定义dp[i][j]为用前j个数取得i个区间，并且第j个数包含在最后一个区间时的最优解，则有递推式：dp[i][[j]=max(dp[i][j-1]+a[j], dp[i-1][t]+a[j])，即这里有两种情况，

一种是直接在前j-1个数形成i个区间末尾加上a[j]，因为dp[i][j-1]的最后一个区间包含a[j-1]，所以这种情况区间数是不增加的。

第二种情况，是在形成i-1个区间的情况下，以a[j]为一个新区间，这样就有了i个区间，这里i-1<=t<=j-1。

但是如果这么做复杂度会达到O(n^3),需要降低复杂度。

我们发现，dp[i-1][t]最大值可以同样的以dp的思路记录下来，保存在M[]数组内，每次求dp[i][j]时不用在[i-1, j-1]枚举dp[i-1][t]直接拿M[j-1]就行了。

优化一下代码，dp[i-2][j]在后面的计算用不到了，dp[i-1][j]的最大值被存了下来，于是可以把dp[i][j]改成dp[i]节省空间。

 

AC代码：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 const int MAXN=1e6+10;
 6 const long long INF=1e12+10;
 7 long long dp[MAXN];
 8 long long a[MAXN],M[MAXN];
 9 int main()
10 {
11     int k,n;
12     while(~scanf("%d %d", &k, &n))
13     {
14         for(int i=1;i<=n;i++){
15             scanf("%lld", &a[i]);
16         }
17         memset(dp, 0, sizeof(dp));
18         memset(M, 0, sizeof(M));
19         long long res;
20         for(int i=1;i<=k;i++){
21             res=-INF;
22             for(int j=i;j<=n;j++){
23                 dp[j]=max(dp[j-1], M[j-1])+a[j];
24                 M[j-1]=res;
25                 res=max(res, dp[j]);
26             }
27         }
28         printf("%d\n", res);
29             
30     }
31 } 

 

转载于:https://www.cnblogs.com/MasterSpark/p/7517715.html