[kuangbin带你飞]专题1-23 Max Sum Plus Plus HDU - 1024

总结：DP思想

原题链接：https://vjudge.net/problem/HDU-1024

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

思路讲解：一开始会又想要采用二维数组的写法采用dp[i][j]来进行存储前i个数分为j块的情况
          那么将分为dp[i][j]将取三个值中的最大值  dp[i-1][j]+a[i](加载最后一组里）   dp[i-1][j-1]（舍弃掉这个数）   dp[k][j-1]+a[i]（新开一组）
          但是由于题目达到1000006 数组无法放下 所以开一个数组a[j]并开两个循环来进行模拟
          具体看注释

 ac代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<algorithm>
#include<stdio.h>
using namespace std;
int a[1000005],dp[1000005],temp[200005];
int main()
{
    int n,m,i,j;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(temp,0,sizeof(temp));
        memset(a,0,sizeof(a));
        for(i=1;i<=n;i++)//输入 
            scanf("%d",&a[i]);
        int ans;
        for(i=1;i<=m;i++)//i来代表分为多少块 从1块开始 
        {
            ans=-99999999;//这里最开始是为了将每次划分的最开始设为极小值在后面更新划分 
            for(j=i;j<=n;j++)//j表示前j个数分为i块的最大值 j应当大于i 
            {
                //dp表示分为i块前j数的最大值   temp为分为i-1块前j数最大值 
                dp[j]=max(dp[j-1]+a[j],temp[j-1]+a[j],);
                temp[j-1]=ans;
                ans=max(ans,dp[j]);//与不加j的值进行比较 并对此时的最优值进行赋值 
            }
        }
        cout<<ans<<endl;
    }
}