hdu 1024 最大上升子序列

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31477    Accepted Submission(s): 11131

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.

題意就是求一段序列中給定條數的子序列的最大總值
那麼就有在遍歷該序列的每一個數來選子序列時就有兩種方法：
1.把該數作爲單獨的一段
2.把該數與前一個數所在的子序列融合
所以，狀態轉移方程就是：
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j - 1]) + str[j];
其中，i 表示段數，j 表示第 j 個數，str【】表示總序列

代碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
typedef long long LL;
///o(っ。Д。)っ AC萬歲!!!!!!!!!!!!!!


const int maxn = 1700000, inf = -199999999;
LL str[maxn] = {}, dp[2][maxn] = {};


int main()
{
    int n, m;
    while(~scanf("%d %d", &m, &n))
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%lld", &str[i]);
            dp[1][i] = dp[0][i] = 0;
        }
        dp[1][0] = dp[0][0] = 0;
        LL maxx = inf;
        for(int i = 1; i <= m; i++)
        {
            maxx = inf;
            for(int j = i; j <= n; j++)
            {
                dp[1][j] = max(dp[1][j - 1], dp[0][j - 1]) + str[j];
                dp[0][j - 1] = maxx;
                maxx = max(maxx, dp[1][j]);
            }
        }
        printf("%lld\n", maxx);
    }
    return 0;
}
/*
3 5
1 -2 -3 0 -1
2 5
1 -2 -3 0 -1
*/