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  • HDU-1024_Max Sum Plus Plus

    Max Sum Plus Plus

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

    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.

    题目大意:给N个数字,然后取X段的区间,要求这些区间的和最大。注意:区间不能重合。

    题解:用f[i][j]表示前i个数第j个不相交的子段的最大值,则f[i][j] = max(f[i]][j],f[i][j-1]+a[i])。但是数组是1e6,会炸。所以用滚动数组,以j为最外层循环,这样f[i][j-1]可以在上一层处理出来。

    附上代码

    #include <algorithm>
    #include <iostream>
    #include <cstdlib>
    #include <cstring>
    #include <cstdio>
    #include <queue>
    #include <stack>
    
    using namespace std;
    const int INF = 1e9+7;
    const int maxn = 1000050;
    
    int a[maxn],dp[maxn],Max[maxn];//dp记录f[i][j-1],Max记录前i个f[i][j-1]最大值。
    
    int main()
    {
        int m,n,i,j,Max_sum;
        while(scanf("%d%d",&m,&n)!=EOF)
        {
            for(i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
                dp[i] = Max[i] = 0;
            }
            for(i=1;i<=m;i++)
            {
                Max_sum = -INF;
                for(j=i;j<=n;j++)
                {
                    dp[j] = a[j] + max(dp[j-1],Max[j-1]);
                    Max[j-1] = Max_sum;
                    Max_sum = max(Max_sum,dp[j]);
                }
            }
            printf("%d
    ",Max_sum);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/luoxiaoyi/p/9705581.html
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