hdu 1024 dp 最长连续子序列和加强版

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.

dp[i][j]表示在有i段子序列的时候，前j个元素的最大和

dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j])要么是直接加入第i段的和，要么是令开一段

这里有时间和空间的复杂度问题一定要注意，这里maxx不仅是求出当前的最大值，其实还充当了一个临时变量的作用，因为当我们使用了pre[j-1]的时候，那么pre[j-1]是不是就没有用处了，所以我们直接将maxx的值赋值覆盖，然后maxx其实更新的是这时候的最大值，这就是和max(dp[i-1][k])相当，因为这是在第i段时候前j个元素的最大值，那么我们在第i+1段的时候，去使用它那么就是对的。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
int dp[1000009];
int a[1000009];
int pre[1000009];
int main()
{
    int n,m,i,j,k,maxx;
    while(~scanf("%d%d",&m,&n))
    {
        memset(dp,0,sizeof(dp));
        memset(pre,0,sizeof(pre));
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        //dp[i][j]表示在前j个元素，取i段情况
        //分两种情况，是第i组接在前面，还是自成一段
        for(i=1; i<=m; i++)
        {
            maxx=-99999999;
            for(j=i; j<=n; j++)
            {
                dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]);
                //dp[j-1]+a[j]相当于dp[i][j-1]+a[j]在后面继续接
                //pre[j-1]+a[j]相当于max(dp[i-1][k])+a[j],即自成一段
                pre[j-1]=maxx;
                maxx=max(dp[j],maxx);
                //pre[j-1]也就是在k=(i,j)情况下，dp[i-1][k]的最大值，这样降低时空复杂度

            }
        }
        printf("%d\n",maxx);
    }

}

hdu 1024 dp 最长连续子序列和加强版

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