2014-04-29 04:36
题目:最大子数组和的二位扩展:最大子矩阵和。
解法:一个维度上进行枚举,复杂度O(n^2);另一个维度执行最大子数组和算法,复杂度O(n)。总体时间复杂度为O(n^3),还需要O(n)额外空间。
代码:
1 // 18.12 Given an n x n matrix, find the submatrix with largest sum. Return the sum as the result. 2 #include <algorithm> 3 #include <climits> 4 #include <iostream> 5 #include <vector> 6 using namespace std; 7 8 class Solution { 9 public: 10 int largestSubmatrixSum (const vector<vector<int> > &matrix) { 11 n = matrix.size(); 12 if (n == 0) { 13 return 0; 14 } 15 m = matrix[0].size(); 16 if (m == 0) { 17 return 0; 18 } 19 20 int i, j, k; 21 vector<int> v; 22 int msum; 23 int sum; 24 25 v.resize(m); 26 msum = INT_MIN; 27 for (i = 0; i < n; ++i) { 28 fill(v.begin(), v.end(), 0); 29 for (j = i; j < n; ++j) { 30 for (k = 0; k < m; ++k) { 31 v[k] += matrix[j][k]; 32 } 33 sum = maxSubarraySum(v, m); 34 msum = max(msum, sum); 35 } 36 } 37 v.clear(); 38 return msum; 39 }; 40 private: 41 int n, m; 42 43 int maxSubarraySum(const vector<int> &v, int n) { 44 int msum; 45 int sum; 46 int i; 47 48 msum = INT_MIN; 49 for (i = 0; i < n; ++i) { 50 if (v[i] >= 0) { 51 msum = max(msum, v[i]); 52 break; 53 } 54 } 55 if (i == n) { 56 return msum; 57 } 58 59 msum = sum = 0; 60 for (i = 0; i < n; ++i) { 61 sum += v[i]; 62 msum = max(msum, sum); 63 sum = max(sum, 0); 64 } 65 66 return msum; 67 }; 68 }; 69 70 int main() 71 { 72 int i, j; 73 int n, m; 74 vector<vector<int> > matrix; 75 Solution sol; 76 77 while (cin >> n >> m && (n > 0 && m > 0)) { 78 matrix.resize(n); 79 for (i = 0; i < n; ++i) { 80 matrix[i].resize(m); 81 for (j = 0; j < m; ++j) { 82 cin >> matrix[i][j]; 83 } 84 } 85 cout << sol.largestSubmatrixSum(matrix) << endl; 86 87 for (i = 0; i < n; ++i) { 88 matrix[i].clear(); 89 } 90 matrix.clear(); 91 } 92 93 return 0; 94 }