zoj 1017

本题的思路在于将组成多边形的每一个最小的三角形都“浓缩”为一个点。然后将正放的小三角形所对应的坐标的y值设为偶数，倒放的为基数，再用手头已有的三角形去覆盖，按行覆盖即可

// C++Exercise.cpp : 定义控制台应用程序的入口点。//#include<iostream>#include<string>#include<cstring>#include<stack>#include<map>#include<vector>#include<algorithm>#include<iomanip>#include<cmath>#include<cstdio>#include<sstream>using namespace std;int t, s, n;bool area[110][110];bool feasible(int x, int y, int length){if (x + length - 1 > 2 * s || y + 2 * length - 2 > 4 * s) return false;if (y % 2 == 1){//上三角for (int i = 0; i < length; i++){for (int j = 0; j < 2 * i + 1; j++){if (area[x + i][y + j] == false) return false;}}}else{//下三角for (int i = 0; i < length; i++){for (int j = 2 * i; j < 2 * length - 1; j++){if (area[x + i][y + j] == false) return false;}}}return true;}void change(int x, int y, int length){if (y % 2 == 1){//上三角for (int i = 0; i < length; i++){for (int j = 0; j < 2 * i + 1; j++){area[x + i][y + j] = false;}}}else{//下三角for (int i = 0; i < length; i++){for (int j = 2 * i; j < 2 * length - 1; j++){area[x + i][y + j] = false;}}}}void restore(int x, int y, int length){if (y % 2 == 1){//上三角for (int i = 0; i < length; i++){for (int j = 0; j < 2 * i + 1; j++){area[x + i][y + j] = true;}}}else{//下三角for (int i = 0; i < length; i++){for (int j = 2 * i; j < 2 * length - 1; j++){area[x + i][y + j] = true;}}}}bool dfs(int x, int y, vector<int>& small){if (x > 2 * s) return true;if (y > 4 * s) return dfs(x + 1, 1, small);if (area[x][y] == false){int j;for (j = y + 1; j <= 4 * s; j++){if (area[x][j] == true) break;}return dfs(x, j, small);}for (int i = 0; i < n; i++){if (feasible(x, y, small[i])){change(x, y, small[i]);if (dfs(x, y + 1, small)) return true;restore(x, y, small[i]);}else break;}return false;}int main(){cin >> t;while (t){cin >> s >> n;vector<int> small;for (int i = 0; i < n; i++){int t1;cin >> t1;small.push_back(t1);}bool success = false;;sort(small.begin(), small.end());for (int i = 0; i < n; i++){if (small[i] > s){n = i;break;}}for (int i = 0; i < n; i++){if (s%small[i] == 0){success = true;break;}else{for (int j = 0; j < i; j++){if (small[i] % small[j] == 0){for (int k = i + 1; k < n; k++) small[k - 1] = small[k];n--;i--;break;}}}}if (success){cout << "YES" << endl;t--;continue;}memset(area, false, sizeof(area));for (int i = 1; i <= s; i++){//上三角开始for (int j = 1; j <= 2 * s + 2 * i - 1; j++){area[i][j] = true;}}for (int i = s + 1; i <= 2 * s; i++){//下三角开始for (int j = 2 * i - 2 * s; j <= 4 * s; j++){area[i][j] = true;}}if (dfs(1, 1, small)) cout << "YES" << endl;else cout << "NO" << endl;t--;}return 0;}