Ma moved to B shape the rules of chess in China.
Please write a program, a given board size n * m, and the initial position of the horse (x, y), a request can not be repeated through the same point on the board, the horse can calculate how many routes through all points on the board.
Input first line integer T (T <10), indicates the number of test data sets.
Each set of test data contains a line of four integers, respectively, the board size and an initial position coordinate n, m, x, y. (0 <= x <= n -1,0 <= y <= m-1, m <10, n <10) Output test data comprising each line, is an integer, the total number of horses can traverse the route board representation, 0 can not be traversed once. Sample Input
1 5 4 0 0
Sample Output
32
#include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> using namespace std; #define ull unsigned long long #define lli long long #define pq priority_queue<int> #define pql priority_queue<ll> #define pqn priority_queue<node> #define v vector<int> #define vl vector<ll> #define read(x) scanf("%d",&x) #define lread(x) scanf("%lld",&x); #define pt(x) printf("%d\n",(x)) #define yes printf("YES\n"); #define no printf("NO\n"); #define gcd __gcd #define out(x) cout<<x<<endl; #define over cout<<endl; #define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++) #define input(k) for (int i = 1; i <= (int)(k); i++) {scanf("%d",&a[i]) ; } #define mem(s,t) memset(s,t,sizeof(s)) #define ok return 0; #define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); #define mod(x) ((x)%9973) #define test cout<<" ++++++ "<<endl; //二叉树 #define lson rt<<1, l, m #define rson rt<<1|1, m+1, r //线段树 #define ls now<<1 #define rs now<<1|1 //int dir[4][2] = {1,0,-1,0,0,1,0,-1}; //单位移动 int dir[8][2] = {2,1,2,-1,-2,1,-2,-1,1,2,1,-2,-1,2,-1,-2}; int t,n,m,k,x,y,col,ex,ey,ans,cnt; int a[202][202],b[202][202]; int vis[11][11]; struct node { int x,y,t;}; void DFS(int r,int c) { if(ans>=n*m) {cnt++;return ;} for(int i=0;i<8;i++) { int nx = r + dir[i][0]; int ny = c + dir[i][1]; if(nx<0 || ny<0 ||nx >=n ||ny >=m) continue; if(!vis[nx][ny]) { vis[nx][ny] = 1; ans++; DFS(nx,ny); ans--; vis[nx][ny] = 0; } } } int main() { cin>>t; while(t--) { cin>>n>>m>>x>>y; for(int i=0;i<n;i++) for(int j=0;j<m;j++) vis[i][j]=0; vis[x][y] = 1; ans=1;cnt=0; DFS(x,y); cout<<cnt<<endl; } ok; }