UVa Problem 108 – Maximum Sum

// UVa Problem 108 - Maximum Sum// Verdict: Accepted// Submission Date: 2011-11-23// UVa Run Time: 0.112s//// 著作權（C）2011，邱秋。metaphysis # yeah dot net//// [解題方法]// 該題可以歸結為網格上的演算法。實際上可以通過正交範圍查詢來解決，若是用嵌套迴圈來求矩形內元素和，難// 免 TLE。可以參考《挑戰編程：程式設計競賽訓練手冊》第十四章第五節的內容。#include <iostream>#include <cstring>#include <ctime>using namespace std;#define MAXN 110int grid[MAXN][MAXN], dominance[MAXN][MAXN], n;int main (int argc, char const* argv[]){// time_t start = clock();while (cin >> n){for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)cin >> grid[i][j];// 行優先建立優勢矩陣。memset(dominance, 0, sizeof(dominance));for (int i = 0; i < n; i++)for (int j = 0; j < n; j++){dominance[i][j] += grid[i][j];if (i > 0)dominance[i][j] += dominance[i - 1][j];if (j > 0)dominance[i][j] += dominance[i][j - 1];if (i > 0 && j > 0)dominance[i][j] -= dominance[i - 1][j - 1];}// 利用正交範圍查詢找最大和。int maxSum = 0;for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)for (int k = i; k < n; k++)for (int l = j; l < n; l++){int tmpSum = dominance[k][l];if (i > 0)tmpSum -= dominance[i - 1][l];if (j > 0)tmpSum -= dominance[k][j - 1];if (i > 0 && j > 0)tmpSum += dominance[i - 1][j - 1];maxSum = max(maxSum, tmpSum);}cout << maxSum << endl;}// cout << (clock() - start) << endl;return 0;}