[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;
else
return 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;
}