hdoj problem 1024Max Sum Plus Plus（动态规划&&DP问题）

最新推荐文章于 2020-09-26 22:19:04 发布

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最新推荐文章于 2020-09-26 22:19:04 发布

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分类专栏：动态规划 DP hdoj 代码 ACM模板

本文链接：https://blog.youkuaiyun.com/lh__huahuan/article/details/40626213

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本文介绍了一个名为 MaxSumPlusPlus 的问题，该问题要求在给定序列中找出 m 对下标，使得这些下标的子序列和之总和最大。文章通过动态规划的方法解决了这一挑战，并给出了具体的实现代码。

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17208 Accepted Submission(s): 5669

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 ... 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(i₁, j₁) + sum(i₂, j₂) + sum(i₃, j₃) + ... + 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 S₁, S₂, S₃ ... 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.

Author

JGShining（极光炫影）

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/*
dp思想
dp[i][j]=Max(dp[i][j-1]+pre[i-1][k])+a[j];//条件j<=i+n-m;
pre[i-1][k]之前最优解法；
dp[i][j]当前最优解法；保存当前状态
a[j]结束状态；
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define M 1000001
int data[M];
int pre[M];
int dp[M];
int Max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int m,n;
while(~scanf("%d%d",&m,&n))
{
int i,j,k;
int max;

for(i=1;i<=n;i++)
scanf("%d",&data[i]);
memset(dp,0,sizeof(dp));
memset(pre,0,sizeof(pre));

if(n<m||m<1)
max=0;
dp[0]=0;
pre[1]=0;
for(i=1;i<=m;i++)
{
dp[i]=dp[i-1]+data[i];
pre[i-1]=dp[i];

max=dp[i];
for(j=i+1;j<=i+n-m;j++)
{
dp[j]=Max(dp[j-1],pre[j-1])+data[j];
pre[j-1]=max;//这两条语句不能写反了,pre[j-1]表示的是上一个状态中i...j-1的最大值，max更新之后表示的i...j的最大值，

max=Max(max,dp[j]);
}
pre[j-1]=max;
}

for(i=m;i<=n;i++)
if(max<dp[i])
max=dp[i];

printf("%d\n",max);
}
return 0;
}