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以下实现最大子数组的分治策略，算法来自《算法导论》
#include<iostream>
using namespace std;
struct ans
{
int low;
int high;
int sum;
};
ans MAXIMUM_CROSSING_SBUARRAY(int * A, int low, int mid, int high)
{
ans cross_ans = {mid, mid, INT_MIN};
ans cross_left = {mid, mid, INT_MIN};
ans cross_right = {mid + 1, mid + 1, INT_MIN};
int sum = 0;
for(int i = mid; i >= low; --i)
{
sum += A[i];
if(cross_left.sum < sum)
{
cross_left.sum = sum;
cross_left.low = i;
}
}
sum = 0;
for (int j = mid + 1; j <= high; ++j)
{
sum += A[j];
if(cross_right.sum < sum)
{
cross_right.sum = sum;
cross_right.high = j;
}
}
cross_ans.low = cross_left.low;
cross_ans.high = cross_right.high;
cross_ans.sum = cross_left.sum + cross_right.sum;
return cross_ans;
}
ans MAXIMUM_SUBARRAY(int * A, int low, int high)
{
if(low == high)
{
ans myans = {low, high, A[low]};
return myans;
}
else
{
int mid = (low + high) / 2;
ans leftans = MAXIMUM_SUBARRAY(A, low, mid);
ans rightans = MAXIMUM_SUBARRAY(A, mid + 1, high);
ans crossans = MAXIMUM_CROSSING_SBUARRAY(A, low, mid, high);
if(leftans.sum > rightans.sum && leftans.sum > crossans.sum)
return leftans;
else if(rightans.sum > leftans.sum && rightans.sum > crossans.sum)
return rightans;
else
return crossans;
}
}
int main()
{
int b[] = {3, -2, 5, -7, 3, 1, 1,-4, 9, -3};
ans mybestans = MAXIMUM_SUBARRAY(b, 0, 9);
cout << "The sub_array‘s elements of b with the greatest sum is the element from "<< mybestans.low << " to " << mybestans.high << endl;
cout << "The sum is " << mybestans.sum << endl;
return 0;
}
最大子数组分治方案C++实现
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原文地址：http://www.cnblogs.com/xiaoyajiang/p/5950201.html
