HDOJ 1024 Max Sum Plus Plus -- 动态规划

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

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 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

   
      
  
    

 

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分析
设状态为 cur[i,j]，表示前 j 项分为 i 段的最大和，且第 i 段必须包含 data[j]，则状态转移方程如下：
cur[i,j] = max{cur[i,j − 1] + data[j],max{cur[i − 1,t] + data[j]}}, 其中i ≤ j ≤ n,i − 1 ≤ t < j
target = max{cur[m,j]}, 其中m ≤ j ≤ n
分为两种情况：
• 情况一，data[j] 包含在第 i 段之中，cur[i,j − 1] + data[j]。
• 情况二，data[j] 独立划分成为一段，max{cur[i − 1,t] + data[j]}。
观察上述两种情况可知 cur[i,j] 的值只和 cur[i,j-1] 和 cur[i-1,t] 这两个值相关，因此不需要二维数组，
可以用滚动数组，只需要两个一维数组，用 cur[j] 表示现阶段的最大值，即 cur[i,j − 1] + data[j]，用
pre[j] 表示上一阶段的最大值，即 max{cur[i − 1,t] + data[j]}。

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int MaxSum(int * data, int m, int n){
    int i, j, max_sum;
    int * cur = (int *)calloc(n + 1, sizeof(int));
    int * pre = (int *)calloc(n + 1, sizeof(int));
    data = data - 1;  //data下标从0开始， cur、pre下标从1开始，为使下标一致，data减1
    for (i = 1; i <= m; ++i){
        max_sum = INT_MIN;
        for (j = i; j <= n; ++j){
            if (cur[j - 1] < pre[j - 1])
                cur[j] = pre[j - 1] + data[j];
            else
                cur[j] = cur[j - 1] + data[j];
            pre[j - 1] = max_sum;
            if (max_sum < cur[j])
                max_sum = cur[j];
        }
        pre[j - 1] = max_sum;
    }
    free(cur);
    free(pre);
    return max_sum;
}

int main(void){
    int m, n, i, *data;
    while (scanf("%d%d", &m, &n) != EOF){
        data = (int *)malloc(sizeof(int) * n);
        for (i=0; i<n; ++i){
            scanf("%d", &data[i]);
        }
        printf ("%d\n", MaxSum(data, m, n));
        free(data);
    }

    return 0;
}

参考资料： ACM Cheat Sheet