• HDU 1024 Max Sum Plus Plus


    题目链接: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): 21926    Accepted Submission(s): 7342


    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)
     
    题意:输入m,n,然后输入n个数。求最大连续m段和。
     
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    
    using namespace std;
    
    const int oo = 0x7fffffff;
    int a[1000005];
    int dp[1000005];
    int Max[1000005];
    
    int main()
    {
        int m,n;
        while(~scanf("%d%d", &m, &n))
        {
            for(int i=1; i<=n; i++)
            {
                scanf("%d", &a[i]);
                Max[i] = 0;
                dp[i] = 0;
            }
            int M;
            dp[0] = 0;
            Max[0] = 0;
            for(int i=1; i<=m; i++)
            {
                M = -oo;
                for(int j=i; j<=n; j++)
                {
                    dp[j] = max(dp[j-1]+a[j], Max[j-1]+a[j]);//其中dp[j-1]表示的是以j-1结尾的元素i个子段的数和,Max[j-1]表示的是前j-1个元素中i-1个子段的数和
                    Max[j-1] = M;//更新Max数组,下次循环用到。放在此处是为了实现Max[j-1]+a[j]中a[j]是一个独立的子段,那么此时就应该用的是i-1段
                    M = max(dp[j], M);//更新M
                }
            }
            printf("%d
    ",M);
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/mengzhong/p/5058186.html
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