HDU 1024 Max Sum Plus Plus(m个子段的最大子段和)

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35988    Accepted Submission(s): 12807


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)
 
分析:
m个子段的最大子段和
子段不重合
第一次写这个问题,还不是很懂
我是参考的大佬的博客:
 
回来更新一波
今天训练赛又遇到这个问题了,说说我的理解
这个问题的dp做其实跟LIS和LCS的有一丁点的像的
外层n循环,i
内层 0到i-1循环 j
每第i次选择都是在前面的i个中(不包括i)选最大的加上当前a【i】的值
上面都是自己瞎逼逼,不好理解,看下面
 
分析:
最大m段字段和问题是由最大字段和问题演变来的
最大字段和问题大家应该都知道
最大字段和是把每个数据看成一个对象,从而组成一个子段
 
核心:那么最大m段字段和,就是把一个子段看成一个对象,再进行最大字段和的操作,从而组成m个最大子段,所以是对象变了(自己也是突然想到)
 
 
code:
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31};
int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

#define INF 0x7fffffff//无穷大
#define max_v 1000010
int a[max_v];
int now[max_v];// now[j]:包含第j个元素的最大和
int pre[max_v];// pre[j]:前j个元素的最大和,不包括第j个元素
int main()
{
    int n,m,maxx;
    while(~scanf("%d %d",&m,&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(now,0,sizeof(now));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=m;i++)//m个子段
        {
            maxx=-INF;
            for(int j=i;j<=n;j++)
            {
                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个子段后面加a[j],一种是a[j]单独成为一个子段
                pre[j-1]=maxx;//更新pre使得pre[j-1]是前j-1个中的最大子段和
                maxx=max(now[j],maxx);
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}

 

 

 

posted @ 2018-07-15 19:06  西*风  阅读(371)  评论(0编辑  收藏  举报