HDU1024——Max Sum Plus Plus（dp）



Max Sum Plus Plus

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25456    Accepted Submission(s): 8795

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 ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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 S₁, S₂, S₃ ... 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

题意：

给定一个数组，求m个不相交子段最大值。

解题思路：

dp[i][j] = max( dp[i][j-1] + a[j], dp[i-1][j-1] + a[j] ).

dp[i][j]表示数组到第j个数时分成i段最大的值（好好理解），dp[i][j-1]+a[j]表示第j个数包括在第i段的情况，dp[i-1][j-1]+a[j]表示第j个数是第i段的第一个数的情况。

由于题目的要求是1,000,000个数而且每个数又很大，所以用二维数组很快就爆栈了，这里咱们采用两个一维滚动数组，核心思想是一样的。

now[j] 数组表示第j个数包括在第i段 , pre[j]数组表示第j个数是第i段的开头。

核心代码：

for(i=1;i<=m;i++)

 {             Max=-99999999;      for(j=i;j<=n;j++) // 想想j为什么从=i开始      {            now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]);            //now[j]有两种来源，一种是直接在第i个子段之后添加a[j]            //一种是是a[j]单独成为1个子段            pre[j-1]=Max;       //更新pre使得pre是前j-1个中最大子段和            if(now[j]>Max)  Max=now[j];      }

 }

出处

代码实现：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long LL;
const int maxn = 1000000+10;
const int inf = -0x7fffffff;
int a[maxn];
int pre[maxn];
int now[maxn];
int MAX;

int main()
{
    int m, n;
    while( ~scanf("%d%d",&m,&n) )
    {
        memset(a,0,sizeof(a));
        memset(pre,0,sizeof(pre));
        memset(now,0,sizeof(now));
        int i,j;
        for( i=1; i<=n; i++ )
            scanf("%d",&a[i]);
        for( i=1; i<=m; i++ )
        {
            MAX = inf;
            for( j=i; j<=n; j++ )
            {                                              // 顺序不能颠倒，想想颠倒的后果；
                now[j] = max(now[j-1]+a[j], pre[j-1]+a[j]);
                pre[j-1] = MAX;
                MAX = max(MAX, now[j]);
            }
        }
        printf("%d\n",MAX);
    }
    
    return 0;
}