/*　fi表示第i行的最左最小元素的列小標，則有f0<f1<f2<...<fn-1取數組的偶數行，組成新的子數組，遞歸求解最左最小元素的列下表，利用偶數項限定奇數項的范圍，再求奇數項 */ #include<iostream> #include<vector> using namespace std; void findOddMin(vector<vector<int>> &a, int m, int n, vector<int> &r) {int len = r.size();vector<int> tmp;for (int i = 0; i < m; ++i){int b = 0, e = n - 1;if (i % 2 == 0){if (i / 2>0) b = r[i / 2 - 1];if (i / 2 < len) e = r[i / 2];int k = b;for (int j = b + 1; j <= e; ++j)if (a[i][j] < a[i][k]) k = j;tmp.push_back(k);}else tmp.push_back(r[i / 2]);}r = tmp; } void findMin(vector<vector<int>> &a, int m, int n, vector<int> &r) {if (m == 1){int k = 0;for (int i = 1; i < n; ++i)if (a[0][i] < a[0][k]) k = i;r.push_back(k);return;}vector<vector<int>> b;for (int i = 0; i < m; ++i)if (i % 2 == 1) b.push_back(a[i]);findMin(b, m / 2, n, r);findOddMin(a, m, n, r); }int main() {//vector<vector<int>> a = { { 10, 13, 23 }, { 17, 16, 23 }, { 45, 32, 23 } };vector<vector<int>> a = { { 37, 23, 22, 32 }, { 21, 6, 5, 10 }, { 53, 34, 30, 31 }, { 32, 13, 9, 6 }, { 43, 21, 15, 8 } };vector<int> r;findMin(a, 5,4, r);for (auto i : r)cout << i << ' ';cout << endl; }

　　

轉載于:https://www.cnblogs.com/jokoz/p/4693042.html

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