[Topcoder] SRM197

divII lev2

 

先定义八个方向的偏移。取数组pieces第一对坐标，得到其八个偏移坐标，结果肯定是这些坐标的子集，所以对这八个坐标进行检醒，如果坐标对数组其它坐标都形成威胁，则加入结果数组。题目要求排好序，只要定义偏移时按有序排列就行，这样省得显式排序了。

#include <iostream>#include <string>#include <vector>#include <cmath>#include <sstream>#include <iterator>using namespace std;#define ABS(x)( x < 0 ? -x : x)const int xoffset[8] = {-2, -2, -1, -1, 1, 1, 2, 2};const int yoffset[8] = {-1, 1, -2, 2, -2, 2, -1, 1};class GeneralChess {private:int canThreaten(const int curx, const int cury, const int posx, const int posy) {if (ABS(curx - posx) == 1 && ABS(cury - posy) == 2|| ABS(curx- posx) == 2 && ABS(cury - posy) == 1)return 1;else return 0;}public:vector<string> attackPositions(vector<string> pieces) {vector<string> res;int x, y;char t;istringstream ss(pieces.at(0));ss >> x >> t >> y;for (int i = 0; i < 8; i++) {int curx = x + xoffset[i];int cury = y + yoffset[i];int j = 1;for (j = 1; j < pieces.size(); j++) {istringstream oss(pieces.at(j));int posx, posy;oss >> posx >> t >> posy;if (!canThreaten(curx, cury, posx, posy)) {break;}}if (j == pieces.size()) {ostringstream ress;ress << curx << ',' << cury;res.push_back(ress.str());}}return res;}};int main(){GeneralChess gc;vector<string> vec;vec.push_back("0,0");vector<string> res = gc.attackPositions(vec);copy(res.begin(), res.end(), ostream_iterator<string>(cout, " "));cout << endl;vec.clear();vec.push_back("2,1");vec.push_back("-1,-2");res = gc.attackPositions(vec);copy(res.begin(), res.end(), ostream_iterator<string>(cout, " "));cout << endl;return 0;}

当然，如果硬要显式排序，也可以：

bool compare(string a, string b) {int ax, ay;int bx, by;char t;istringstream ss(a);ss >> ax >> t >> ay;ss.clear();ss.str(b);ss >> bx >> t >> by;if (ax < bx || ax == bx && ay < by)return true;elsereturn false;}int main(){vector<string> vec;vec.push_back("0,0");vec.push_back("2,1");vec.push_back("1,2");sort(vec.begin(), vec.end(), compare);copy(vec.begin(), vec.end(),ostream_iterator<string>(cout, " "));cout << endl;return 0;}