hdu1024 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): 21752    Accepted Submission(s): 7282

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.

 

这题m的数据范围不知道，是个坑点，看了别人的博客知道只有n*m的算法才能过= =.

刚开始我的思路是设dp[i][j]表示前i个数取j段且第j段包括第i个数的最大和，考虑最后一个数，这个数可以自成一段或者和之前的数组成一段，得到dp[i][j]=max(dp[k][j]+sum[k'][i]).

但是因为不用每个数都选，和区间dp的做法不同，所以这种方法k要枚举，k'也要枚举，会超时，所以这思路不行。

看了别人的思路，发现可以有另一种dp方法，状态表示和前面一样，就是考虑第i个数是延续前一段还是另起一段，那么 dp[i][j]=max(dp[i-1][j]+num[i],dp[k][j-1]+num[i]) j-1<=k<=i-1 

另外这题因为m的范围不知道，所以要用滚动数组。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 1000000000000000
#define pi acos(-1.0)
#define maxn 1000050
ll num[maxn],sum[maxn];
ll dp[maxn][2];

int main()
{
    ll n,m,i,j,temp,ans;
    while(scanf("%lld%lld",&m,&n)!=EOF)
    {
        sum[0]=0;
        for(i=1;i<=n;i++){
            scanf("%lld",&num[i]);
            sum[i]=sum[i-1]+num[i];
        }
        ans=-inf;
        dp[0][1]=-inf;
        for(i=1;i<=n;i++){
            dp[i][1]=max(num[i],dp[i-1][1]+num[i]  );
            ans=max(ans,dp[i][1]);
        }
        if(m==1){
            printf("%lld\n",ans);continue;
        }

        int pre=1;
        for(j=2;j<=m;j++){
            ans=-inf;
            dp[j][1-pre]=sum[j];
            temp=max(sum[j-1],dp[j][pre]);
            ans=max(ans,dp[j][1-pre]);
            for(i=j+1;i<=n;i++){
                dp[i][1-pre]=max(dp[i-1][1-pre],temp)+num[i]; //temp表示dp[k][j-1]，每次循环就把i的情况添加进去
                temp=max(temp,dp[i][pre]);
                ans=max(ans,dp[i][1-pre]);
            }
            pre=1-pre;
        }
        printf("%lld\n",ans);
    }
    return 0;
}