题目
给定一个残缺的三阶幻方，判断是否只有一组可行解，如果只有一组可行解，打印之。
具体描述请见hihoCoder。
解题思路
对给定的残缺三阶幻方，首先可以通过行列斜之和为15尽可能进行补全；如果只有一种解，那一定可以填满9个数字，除了一种情况：“十字架”形。对“十字架”刑，实际上也只可能有一种解，可以根据已填数字推算出整个幻方。
时间复杂度
为常数。
代码
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| #include <iostream>
#include <vector>
using namespace std;
bool check(int *square) {
int filled = 0;
for (int i = 0; i < 9; i++)
filled += (square[i] != 0);
int lastFilled = 0;
do {
lastFilled = filled;
if (square[4] == 0) {
square[4] = 5;
filled++;
}
int state = (square[0] != 0) + (square[8] != 0);
if (state == 1) {
square[square[0] != 0 ? 8 : 0] = 10 - square[0] - square[8];
filled++;
}
state = (square[1] != 0) + (square[7] != 0);
if (state == 1) {
square[square[1] != 0 ? 7 : 1] = 10 - square[1] - square[7];
filled++;
}
state = (square[2] != 0) + (square[6] != 0);
if (state == 1) {
square[square[2] != 0 ? 6 : 2] = 10 - square[2] - square[6];
filled++;
}
state = (square[3] != 0) + (square[5] != 0);
if (state == 1) {
square[square[3] != 0 ? 5 : 3] = 10 - square[3] - square[5];
filled++;
}
state = (square[0] != 0) + (square[1] != 0) + (square[2] != 0);
if (state == 2) {
square[(square[1] == 0) + (square[2] == 0) * 2] = 15 - square[0] - square[1] - square[2];
filled++;
}
state = (square[2] != 0) + (square[5] != 0) + (square[8] != 0);
if (state == 2) {
square[(square[2] == 0) * 2 + (square[5] == 0) * 5 + (square[8] == 0) * 8] = 15 - square[2] - square[5] - square[8];
filled++;
}
state = (square[8] != 0) + (square[7] != 0) + (square[6] != 0);
if (state == 2) {
square[(square[8] == 0) * 8 + (square[7] == 0) * 7 + (square[6] == 0) * 6] = 15 - square[8] - square[7] - square[6];
filled++;
}
state = (square[6] != 0) + (square[3] != 0) + (square[0] != 0);
if (state == 2) {
square[(square[6] == 0) * 6 + (square[3] == 0) * 3] = 15 - square[6] - square[3] - square[0];
filled++;
}
state = (square[1] != 0) + (square[7] != 0) + (square[3] != 0) + (square[5] != 0) + (square[4] != 0);
int state1 = (square[0] == 0) + (square[2] == 0) + (square[6] == 0) + (square[8] == 0);
if (state == 5 && state1 == 4) {
if ((square[1] == 9 && square[3] == 3) || (square[1] == 3 && square[3] == 9)) {
square[0] = 4;
filled++;
}
else if ((square[1] == 9 && square[3] == 7) || (square[1] == 7 && square[3] == 9)) {
square[0] = 2;
filled++;
}
else if ((square[1] == 1 && square[3] == 3) || (square[1] == 3 && square[3] == 1)) {
square[0] = 8;
filled++;
}
else if ((square[1] == 1 && square[3] == 7) || (square[1] == 7 && square[3] == 1)) {
square[0] = 6;
filled++;
}
}
} while (lastFilled != filled);
return (filled == 9);
}
int main() {
int *square = new int[9];
for (int i = 0; i < 9; i++)
cin >> square[i];
if (check(square))
cout << square[0] << " " << square[1] << " " << square[2] << endl << square[3] << " "
<< square[4] << " " << square[5] << endl << square[6] << " " << square[7] << " "
<< square[8] << endl;
else
cout << "Too Many" << endl;
delete[] square;
return 0;
}
|
