K - Max Sum Plus Plus

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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.

//题意是:第一行 m ，n (n<=1000000) 两个整数，然后第二行 n 个数，求 m 段不相交连续序列最大和。

自己想了一阵，还是没有思路。。。这个题很好。

这篇博客写得十分详细

http://www.cnblogs.com/ACMan/archive/2012/08/09/2630038.html

dp[i][j]代表的状态是 i 段连续序列的最大和，并且最后一段一定包含 num[j]

所以写出状态转移方程 dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]} i-1=<t<j-1

dp[i][j-1]代表连续的状态，后面是不连续的状态，取最大值。

数据很大，只能用一维数组

不断更新状态，用一个数据 big 保存 i 段最大的和，最后直接输出，就是答案

436ms

 1 #include <stdio.h>
 2 #include <string.h>
 3
 4 #define inf 0x7ffffff
 5 #define MAXN 1000005
 6 int num[MAXN];
 7 int dp[MAXN];
 8 int pre[MAXN];
 9
10 int max(int a,int b)
11 {
12     return a>b? a:b;
13 }
14
15 int main()
16 {
17     int m,n;
18     int i,j,big;
19     while (scanf("%d%d",&m,&n)!=EOF)
20     {
21         for (i=1;i<=n;i++)
22         {
23             scanf("%d",&num[i]);
24             pre[i]=0;
25         }
26
27         pre[0]=0;
28         dp[0]=0;
29
30         for (i=1;i<=m;i++)
31         {
32             big=-inf;
33             for (j=i;j<=n;j++)
34             {
35                 dp[j]=max(dp[j-1],pre[j-1])+num[j];//连续的最大和，或者不连续的最大和
36                 pre[j-1]=big;      //保存 i - 1 段 最大和
37                 big=max(big,dp[j]);//保证是 i 段最大的和
38             }
39         }
40         printf("%d\n",big);
41     }
42     return 0;
43 }