smart robot

谭绍晖
2023-12-01

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;
}

 

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