CF-1024-C. Mex in the Grid（蛇形构造转圈矩阵） · znzryb
## 思路讲解
从外往里转会好很多。
## AC代码
[https://codeforces.com/contest/2102/submission/319351416](https://codeforces.com/contest/2102/submission/319351416)
::: 源代码
```cpp
// Problem: C. Mex in the Grid
// Contest: Codeforces - Codeforces Round 1024 (Div. 2)
// URL: https://codeforces.com/contest/2102/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e3)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,K,T;
ll maze[MAXN][MAXN];
ll vis[MAXN][MAXN];
ll dx[4]={0,1,0,-1};
ll dy[4]={1,0,-1,0};
/*
678 0->1->2->3,...,->8
501
432
从外往里转会好很多。
*/
inline void solve(){
cin>>N;
FOR(i,0,N+5){
FOR(j,0,N+5){
maze[i][j]=-1;
}
}
ll idx=N*N-1;
ll x=N,y=N,dir=3;
while(idx>=0){
#ifdef LOCAL
if(idx==1){
debug(dir)
}
#endif
maze[x][y]=idx;
--idx;
#ifdef LOCAL
FOR(i,1,N){
FOR(j,1,N){
cerr<<maze[i][j]<<" ";
}
cerr<<"\n";
}
cerr<<"\n";
#endif
if(x+dx[dir]>N || x+dx[dir]<1){
dir--;
}else if(y+dy[dir]>N || y+dy[dir]<1){
--dir;
}else if(maze[x+dx[dir]][y+dy[dir]]!=-1){
--dir;
}
dir%=4;
if(dir<0) dir+=4;
x+=dx[dir];
y+=dy[dir];
}
FOR(i,1,N){
FOR(j,1,N){
cout<<maze[i][j]<<" ";
}
cout<<"\n";
}
#ifdef LOCAL
FOR(i,0,N+5){
FOR(j,0,N+5){
cerr<<maze[i][j]<<" ";
}
cerr<<"\n";
}
cerr<<"\n";
#endif
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while(T--){
solve();
#ifdef LOCAL
cerr << "-----------" << '\n';
#endif
}
// solve();
return 0;
}
/*
*/
```
:::
## 心路历程（WA，TLE，MLE……）
从里往外转比较烦。
WA
::: 源代码
```cpp
// Problem: C. Mex in the Grid
// Contest: Codeforces - Codeforces Round 1024 (Div. 2)
// URL: https://codeforces.com/contest/2102/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e3)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,K,T;
ll maze[MAXN][MAXN];
ll vis[MAXN][MAXN];
ll dx[4]={0,1,0,-1};
ll dy[4]={1,0,-1,0};
/*
678 0->1->2->3,...,->8
501
432
*/
ll idx=0;
void dfs(ll x,ll y,ll step,ll cnt,ll dir){
if(idx>N*N-1){
return;
}
dir%=4;
if(cnt>3){
cnt=1;
step+=1;
ll dir_1=(dir-1)%4;
if(dir_1<0) dir_1+=4;
maze[x][y]=idx;
++idx;
x+=dx[dir_1];
y+=dy[dir_1];
}
if(dir==0){
FOR(i,y,y+step-2){
maze[x][i]=idx;
++idx;
}
dfs(x,y+step-1,step,cnt+1,dir+1);
}else if(dir==1){
FOR(i,x,x+step-2){
maze[i][y]=idx;
++idx;
}
dfs(x+step-1,y,step,cnt+1,dir+1);
}else if(dir==2){
ROF(i,y,y-step+2){
maze[x][i]=idx;
++idx;
}
dfs(x,y-step+1,step,cnt+1,dir+1);
}else{
ROF(i,x,x-step+2){
maze[i][y]=idx;
++idx;
}
dfs(x-step+1,y,step,cnt+1,dir+1);
}
}
inline void solve(){
cin>>N;
ll n2=ceil(N*1.0/2);
FOR(i,0,N+5){
FOR(j,0,N+5){
maze[i][j]=-1;
vis[i][j]=false;
}
}
idx=0;
dfs(n2,n2,2,1,0);
#ifdef LOCAL
FOR(i,0,N+5){
FOR(j,0,N+5){
cerr<<maze[i][j]<<" ";
}
cerr<<"\n";
}
cerr<<"\n";
#endif
FOR(i,1,N){
FOR(j,1,N){
cout<<maze[i][j]<<" ";
}
cout<<"\n";
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while(T--){
solve();
}
// solve();
return 0;
}
/*
*/
```
:::