题目见：http://noi.openjudge.cn/ch0201/755/
最多翻转15次，所以按次数遍历，对0，1，2…15次中的每一次调用play()进行枚举，注意函数的返回条件以及回溯的处理。
#include <iostream> #include <iomanip> using namespace std; char a[8][8]; int step; bool flag; bool isOver() { char temp = a[0][0]; for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { if (a[i][j] != temp) return false; } } return true; } void change(int m, int n) { if (m < 0 || m > 3 || n < 0 || n > 3) return; if (a[m][n] == 'b') a[m][n] = 'w'; else a[m][n] = 'b'; return; } void flip(int m, int n) { change(m, n); change(m - 1, n); change(m + 1, n); change(m, n - 1); change(m, n + 1); return; } void play(int m, int n, int deep) { if (deep == step) { flag = isOver(); return; } if (flag || m > 3) return; flip(m, n); if (n < 3) play(m, n + 1, deep + 1); else play(m + 1, 0, deep + 1); flip(m, n); if (n < 3) play(m, n + 1, deep); else play(m + 1, 0, deep); return; } int main() { for (int i = 0; i < 4; i++) for (int j = 0; j < 4; j++) cin >> a[i][j]; flag = false; for (step = 0; step < 16; step++) { play(0, 0, 0); if (flag == true) break; } if (flag == true) cout << step << endl; else cout << "Impossible" << endl; //system("pause"); return 0; }