HDOJ 1024 Max Sum Plus Plus -- 动态规划

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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 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 32 6 -1 4 -2 3 -2 3

Sample Output

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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

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