https://ac.nowcoder.com/acm/contest/903/L
题意：给n*n方格，0-9；起点任意，上下左右走，组成一个数值；问最小没出现的数；n<=50;
思路：好像是数值是不会超出1e6.............然后就直接dfs枚举每个数，再找没出现的；
大胆猜！！
#include<algorithm> #include<set> #include<queue> #include<cmath> #include<cstring> #include<iostream> #include<set> #include<vector> #include<queue> #include<cmath> #include<cstdio> #include<map> #include<stack> #include<string> #include<bits/stdc++.h> using namespace std; #define sfi(i) scanf("%d",&i) #define pri(i) printf("%d\n",i) #define sff(i) scanf("%lf",&i) #define ll long long #define ull unsigned long long #define mem(x,y) memset(x,y,sizeof(x)) #define INF 0x3f3f3f3f #define eps 1e-16 #define PI acos(-1) #define lowbit(x) ((x)&(-x)) #define zero(x) (((x)>0?(x):-(x))<eps) #define fl() printf("flag\n") #define MOD(x) ((x%mod)+mod)%mod #define endl '\n' #define pb push_back #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r #define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0) const int maxn=1e6+9; const int mod=1e9+7; inline ll read() { ll f=1,x=0; char ss=getchar(); while(ss<'0'||ss>'9') { if(ss=='-')f=-1;ss=getchar(); } while(ss>='0'&&ss<='9') { x=x*10+ss-'0';ss=getchar(); } return f*x; } int n; int g[58][58]; map<int,bool>num; void dfs(int x,int y,int step,int sum) { if(step<=0) return ; num[sum]=1; if(x-1>=1) dfs(x-1,y,step-1,sum*10+g[x-1][y]); if(x+1<=n) dfs(x+1,y,step-1,sum*10+g[x+1][y]); if(y-1>=1) dfs(x,y-1,step-1,sum*10+g[x][y-1]); if(y+1<=n) dfs(x,y+1,step-1,sum*10+g[x][y+1]); } int main() { //FAST_IO; //freopen("input.txt","r",stdin); cin>>n; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { cin>>g[i][j]; } } for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { dfs(i,j,6,g[i][j]); } } for(int i=0;i<=maxn;i++) { if(!num[i]) { cout<<i<<endl; break; } } return 0; }
