蓝桥杯—剪邮票，对初学算法自己启发挺大的一题

#include <iostream> #include <string.h> using namespace std; int dir[4][2] = {{1,0}, {0,1},{-1,0}, {0,-1}}; int counter; int visited[5][5]; int a[6]; void dfs(int x, int y, int cou); int inde; int count; int main() { for(int i1=1; i1<10; ++i1) for(int i2=i1+1; i2<12; ++i2) for(int i3=i2+1; i3<=12; ++i3) for(int i4=i3+1; i4<=12; ++i4) for(int i5=i4+1; i5<=12; ++i5) { a[1] = i1; a[2] = i2; a[3] = i3; a[4] = i4; a[5] = i5; memset(visited, 0, sizeof(visited)); int x = (i1-1)/4; int y = (i1-1)%4; visited[x][y] = 1; inde = 1; dfs(x, y, 1); // if(inde == 5) // ++count; } cout << counter << '\n'; cout << count; } void dfs(int x, int y, int cou) { if(inde == 5) //116， 从这两个结果看出， 判断条件一定不能放在参数里 ++count; if(cou == 5) ++counter; //49， 从这两个结果看出， 判断条件一定不能放在参数里 else { for(int i=0; i<4; ++i) { int sx = x+dir[i][0]; int sy = y+dir[i][1]; if(sx < 0 || sx > 2 || sy <0 || sy >3) continue; for(int j=1; j<=5; ++j) { int xx = (a[j]-1)/4; int yy = (a[j]-1)%4; if(!visited[sx][sy] && (sx ==xx && sy == yy)) { visited[sx][sy] = 1; ++inde; dfs(sx, sy, cou+1); // visited[sx][sy] = 0; } } } } }