动态规划求不相邻的最大子数组和

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问题：给出一个数组，求出其中一个子集，使得子集中每个元素在原数组中两两都不相邻并使子集的和最大。

思路：先找到最优子结构，对于数组A[0...n]，设r[i][j]为A[i][j]的不相邻最大子数组和，则有

r[i][j] =

max(A[i][j],0) i = j

max(r[i][k-1]+r[k-1][j]) i < j, k = i~j 注意k = i or j时，有点特殊，代码中可以看出来

#include <iostream>#include <stdlib.h>using namespace std;#define MIN -1000000int MaxSum(int *A, int len){int **r = new int* [len];for(int i = 0; i < len; i++)r[i] = new int [len];for (int i = 0; i < len; i++)r[i][i] = A[i] > 0 ? A[i] : 0;int l, i, j, k;for (l = 2; l <= len; l++){for(i = 0; i < len - l + 1; i++){j = l + i - 1;int temp = MIN;for(k = i; k <= j; k++){if (k == i)temp = temp > r[k + 1][j] ? temp : r[k + 1][j];else if (k == j)temp = temp > r[i][k - 1] ? temp : r[i][k - 1];else {int q;q = r[i][k - 1] + r[k + 1][j];temp = temp > q ? temp : q;}}r[i][j] = temp;}}int maxsum = r[0][len - 1];for (int n = 0; n < len; n++)delete[] r[n];delete[] r;return maxsum;}int main(){int A[] = {-1,3,4};int len = sizeof(A) / sizeof(int);cout<<MaxSum(A, len)<<endl;system("pause");}

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