题意：如下的10个格子 
填入0~9的数字。要求：连续的两个数字不能相邻。 （左右、上下、对角都算相邻）
一共有多少种可能的填数方案？
分析：dfs，划定边界，行1~4，列1~3，初始化为INT_INF，这样所填入的数字与INT_INF一定不相邻，所以可不必单独考虑（1,1）和（3，4）两个格子。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[5][5];
int vis[10];
bool judge(int x, int y){
return x >= 1 && x <= 3 && y >= 1 && y <= 4;
}
bool deal(int x, int y, int v){
for(int i = 0; i < 8; ++i){
int tmpx = x + dr[i];
int tmpy = y + dc[i];
if(judge(tmpx, tmpy)){
if(abs(a[tmpx][tmpy] - v) == 1) return false;
}
}
return true;
}
int ans;
void dfs(int x, int y){
if(x == 3 && y == 4){
++ans;
return;
}
for(int i = 0; i <= 9; ++i){
if(!vis[i] && deal(x, y, i)){
vis[i] = 1;
a[x][y] = i;
if(y < 4){
dfs(x, y + 1);
}
else{
dfs(x + 1, 1);
}
vis[i] = 0;
a[x][y] = INT_INF;
}
}
}
int main(){
memset(a, INT_INF, sizeof a);
dfs(1, 2);
printf("%d\n", ans);
return 0;
}