Max Sum Plus Plus

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1024

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

题意

给你n个数，让你选择连续的M段，让你得到最大值

题解：

简单的想一想，dp[i][j]表示，前j个数，我选i段的最大值，那么转移方程,dp[i][j]=max(dp[i][j-1]+a[j],dp[i-1][k]+a[j])

然后滚动数组优化一下

代码：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 1000005

#define mod 10007

#define eps 1e-9

int Num;

char CH[];

//const int inf=0x7fffffff;   //нчоч╢С

const int inf=0x3f3f3f3f;

/*

inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    puts("");

}

*/

//**************************************************************************************

inline int read()

{

    int x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=;if(!x){putchar('');puts("");return;}

    while(x>)CH[++Num]=x%,x/=;

    while(Num)putchar(CH[Num--]+);

    puts("");

}

int dp[maxn+];

int mmax[maxn+];

int a[maxn+];

int main()

{

    int n,m;

    int i,j,mmmax;

    while(scanf("%d%d",&m,&n)!=EOF)

    {

        for(i=;i<=n;i++)

        {

            scanf("%d",&a[i]);

            mmax[i]=;

            dp[i]=;

        }

        dp[]=;

        mmax[]=;

        for(i=;i<=m;i++)

        {

                mmmax=-inf;

                for(j=i;j<=n;j++)

                {

                    dp[j]=max(dp[j-]+a[j],mmax[j-]+a[j]);

                    mmax[j-]=mmmax;

                    mmmax=max(mmmax,dp[j]);

                }

        }

        printf("%d\n",mmmax);

    }

    return ;

}

秒客网

hdu 1024 Max Sum Plus Plus DP