HDU 1024 Max Sum Plus Plus

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16803    Accepted Submission(s): 5525

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_xor 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 32 6 -1 4 -2 3 -2 3

 

Sample Output

68
Hint
Huge input, scanf and dynamic programming is recommended.
 

 

Author

JGShining（极光炫影）

解题思路：求n个数的m个不相交子区间的和的最大值，令dp[i][j]表示以第i个数为结尾的j个区间的最大值，

dp[i][j]=max(dp[i-1][j]+arr[i],max(dp[k][j-1])+arr[i],0<k<i)；

注意到dp[i][j]只与dp[i-1][j]和dp[k][j-1]有关，采用滚动数组优化

同时dp[k][j-1]可以在之前的0---i-1中计算出来用数组保存下

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <vector>#include <bitset>#include <cmath>#include <utility>#define Maxn 1000005#define Maxm 1000005#define lowbit(x) x&(-x)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define PI acos(-1.0)#define make_pair MP#define LL long long #define Inf (1LL<<62)#define inf 0x3f3f3f3f#define re freopen("in.txt","r",stdin)#define wr freopen("out.txt","w",stdout)using namespace std;int arr[Maxn];int dp[Maxn];int dpp[Maxn];int read(){    int res=0;    bool flag=true;    while(1)    {         char ch=getchar();         if(ch>='0'&&ch<='9')             res=res*10+ch-'0';         else if(ch=='-')         {            flag=false;            continue;         }         else             break;    }    return flag?res:-res;}int main(){    int n,m;    int ans;    //re;wr;    while(~scanf("%d%d",&m,&n))    {        getchar();        for(int i=0;i<=n;i++)            arr[i]=dp[i]=dpp[i]=0;        for(int i=1;i<=n;i++)            arr[i]=read();        for(int j=1;j<=m;j++)        {            ans=-inf;            for(int i=j;i<=n;i++)                //下标要从j开始，这样才能保证滚动数组是正确的            {                dp[i]=max(dpp[i-1],dp[i-1])+arr[i];//dp[i][j]=max(dp[i-1][j]+arr[i],max(dp[k][j-1])+arr[i],0<k<i)                dpp[i-1]=ans;//dpp存储max(dp[k][j-1]),0<k<i                ans=max(ans,dp[i]);//ans表示以i为结尾的选择了j段的最大值            }        }        printf("%d\n",ans);    }    return 0;}