USACO Section 2.2 Party Lamps

/*ID: lucien23PROG: lampsLANG: C++*//* * 此题的技巧之处就是需要注意到任何button只要按下2的倍数次就相当于没有按 * 所以其实只需要考虑4个按钮，每个按钮是否被有效按下过一次就好 * 直接使用枚举法，一共只有2^4=16种情况 * 对于每种情况需要知道被按下的有效次数（也就是被按下过的按钮数），必须满足 * (C-有效次数)%2=0才行，这样其他次数才能视为无效 * 然后验证各种情况是否符合要求，将符合要求的情况按序输出即可 */#include <iostream>#include <fstream>#include <vector>#include <algorithm>using namespace std;bool compPro(int *strNum1, int *strNum2){int i = 0;while (1){if (strNum1[i] < strNum2[i])return true;else if (strNum1[i] > strNum2[i])return false;i++;}}int main(){ifstream infile("lamps.in");ofstream outfile("lamps.out");if(!infile || !outfile){cout << "file operation failure!" << endl;return -1;}int N, C;vector<int> vecOnLamps, vecOffLamps;vector<int*> results;infile >> N >> C;int num;infile >> num;while (num != -1){vecOnLamps.push_back(num);infile >> num;}infile >> num;while (num != -1){vecOffLamps.push_back(num);infile >> num;}int count = 0;for (int i=0; i<16; i++){int *lamps = new int[N];for (int j=0; j<N; j++){lamps[j] = 1;}int tempCnt = 0;for (int j=0; j<4; j++){int temp = 1 << j;if ((temp & i) == temp)tempCnt++;}if (C < tempCnt || (C - tempCnt) % 2 != 0){delete[] lamps;continue;}if (((1 << 0) & i) == (1 << 0)){for (int j=0; j<N; j++){if (lamps[j] == 0)lamps[j] = 1;elselamps[j] = 0;}}if (((1 << 1) & i) == (1 << 1)){for (int j=0; j<N; j++){if ((j+1)%2 == 1){if (lamps[j] == 0)lamps[j] = 1;elselamps[j] = 0;}}}if (((1 << 2) & i) == (1 << 2)){for (int j=0; j<N; j++){if ((j+1)%2 == 0){if (lamps[j] == 0)lamps[j] = 1;elselamps[j] = 0;}}}if (((1 << 3) & i) == (1 << 3)){for (int j=0; j<N; j++){if ((j+1)%3 == 1){if (lamps[j] == 0)lamps[j] = 1;elselamps[j] = 0;}}}bool flag = true;for (vector<int>::iterator it=vecOnLamps.begin(); it!=vecOnLamps.end(); it++){if (lamps[*it-1] != 1){flag = false;break;}}if (!flag){delete[] lamps;continue;}for (vector<int>::iterator it=vecOffLamps.begin(); it!=vecOffLamps.end(); it++){if (lamps[*it-1] != 0){flag = false;break;}}if (flag){count++;results.push_back(lamps);} else {delete[] lamps;}}if (count == 0){outfile << "IMPOSSIBLE" << endl;} else {sort(results.begin(), results.end(), compPro);for (int i=0; i<count; i++){for (int j=0; j<N; j++){outfile << results[i][j];}outfile << endl;}}return 0;}