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 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

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[20];
//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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
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("");
}

int dp[maxn+10];
int mmax[maxn+10];
int a[maxn+10];
int main()
{
    int n,m;
    int i,j,mmmax;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            mmax[i]=0;
            dp[i]=0;
        }
        dp[0]=0;
        mmax[0]=0;
        for(i=1;i<=m;i++)
        {
                mmmax=-inf;
                for(j=i;j<=n;j++)
                {
                    dp[j]=max(dp[j-1]+a[j],mmax[j-1]+a[j]);
                    mmax[j-1]=mmmax;
                    mmmax=max(mmmax,dp[j]);
                }
        }
        printf("%d\n",mmmax);

    }
    return 0;
}