Ural 1298 Knight 题解

Ural 1298 Knight 题解

目录

• Ural 1298 Knight 题解
  • 题意
  • 题解
  • 程序

题意

给定一个\(n\times n(1\le n\le8)\)的国际象棋棋盘和一个骑士(基本上相当于中国象棋的马)，问可否用经过每个格子\(1\)次。如果可以，输出路径，否则输出IMPOSSIBLE。

题解

考虑回溯。暴力程序十分好写，但是会超时。

可以用启发式优化。

设当前点为\((x,y)\)，可到达的点为\((x',y')\)。优先选择\((x',y')\)状态种数少的回溯，即可以转移的格子的数量少。

这样优化后就可以过了。

Tip: 优化后很快，为\(0.015s\)。

程序

// #pragma GCC optimize(2)
// #pragma G++ optimize(2)
// #pragma comment(linker,"/STACK:102400000,102400000")
// #include <bits/stdc++.h>
#include <map>
#include <set>
#include <list>
#include <array>
#include <cfenv>
#include <cmath>
#include <ctime>
#include <deque>
#include <mutex>
#include <queue>
#include <ratio>
#include <regex>
#include <stack>
#include <tuple>
#include <atomic>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <chrono>
#include <cstdio>
#include <cwchar>
#include <future>
#include <limits>
#include <locale>
#include <memory>
#include <random>
#include <string>
#include <thread>
#include <vector>
#include <cassert>
#include <climits>
#include <clocale>
#include <complex>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdint>
#include <cstdlib>
#include <cstring>
#include <ctgmath>
#include <cwctype>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <sstream>
#include <ccomplex>
#include <cstdbool>
#include <iostream>
#include <typeinfo>
#include <valarray>
#include <algorithm>
#include <cinttypes>
#include <cstdalign>
#include <stdexcept>
#include <typeindex>
#include <functional>
#include <forward_list>
#include <system_error>
#include <unordered_map>
#include <unordered_set>
#include <scoped_allocator>
#include <condition_variable>
// #include <conio.h>
// #include <windows.h>
using namespace std;
typedef long long LL;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef float fl;
typedef double ld;
typedef long double LD;
typedef pair<int,int> pii;
#if (WIN32) || (WIN64) || (__WIN32) || (__WIN64) || (_WIN32) || (_WIN64) || (WINDOWS)
#define lld "%I64d"
#define llu "%I64u"
#else
#define lld "%lld"
#define llu "%llu"
#endif
#define ui(n) ((unsigned int)(n))
#define LL(n) ((long long)(n))
#define ull(n) ((unsigned long long)(n))
#define fl(n) ((float)(n))
#define ld(n) ((double)(n))
#define LD(n) ((long double)(n))
#define char(n) ((char)(n))
#define Bool(n) ((bool)(n))
#define fixpoint(n) fixed<<setprecision(n)
const int INF=1061109567;
const int NINF=-1044266559;
const LL LINF=4557430888798830399;
const ld eps=1e-15;
#define MOD (1000000007)
#define PI (3.1415926535897932384626433832795028841971)
/*
#define MB_LEN_MAX 5
#define SHRT_MIN (-32768)
#define SHRT_MAX 32767
#define USHRT_MAX 0xffffU
#define INT_MIN (-2147483647 - 1)
#define INT_MAX 2147483647
#define UINT_MAX 0xffffffffU
#define LONG_MIN (-2147483647L - 1)
#define LONG_MAX 2147483647L
#define ULONG_MAX 0xffffffffUL
#define LLONG_MAX 9223372036854775807ll
#define LLONG_MIN (-9223372036854775807ll - 1)
#define ULLONG_MAX 0xffffffffffffffffull
*/
#define MP make_pair
#define MT make_tuple
#define All(a) (a).begin(),(a).end()
#define pall(a) (a).rbegin(),(a).rend()
#define log2(x) log(x)/log(2)
#define Log(x,y) log(x)/log(y)
#define SZ(a) ((int)(a).size())
#define rep(i,n) for(int i=0;i<((int)(n));i++)
#define rep1(i,n) for(int i=1;i<=((int)(n));i++)
#define repa(i,a,n) for(int i=((int)(a));i<((int)(n));i++)
#define repa1(i,a,n) for(int i=((int)(a));i<=((int)(n));i++)
#define repd(i,n) for(int i=((int)(n))-1;i>=0;i--)
#define repd1(i,n) for(int i=((int)(n));i>=1;i--)
#define repda(i,n,a) for(int i=((int)(n));i>((int)(a));i--)
#define repda1(i,n,a) for(int i=((int)(n));i>=((int)(a));i--)
#define FOR(i,a,n,step) for(int i=((int)(a));i<((int)(n));i+=((int)(step)))
#define repv(itr,v) for(__typeof((v).begin()) itr=(v).begin();itr!=(v).end();itr++)
#define repV(i,v) for(auto i:v)
#define repE(i,v) for(auto &i:v)
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x) MS(x,0)
#define MINF(x) MS(x,63)
#define MCP(x,y) memcpy(x,y,sizeof(y))
#define sqr(x) ((x)*(x))
#define UN(v) sort(All(v)),v.erase(unique(All(v)),v.end())
#define filein(x) freopen(x,"r",stdin)
#define fileout(x) freopen(x,"w",stdout)
#define fileio(x)\
	freopen(x".in","r",stdin);\
	freopen(x".out","w",stdout)
#define filein2(filename,name) ifstream name(filename,ios::in)
#define fileout2(filename,name) ofstream name(filename,ios::out)
#define file(filename,name) fstream name(filename,ios::in|ios::out)
#define Pause system("pause")
#define Cls system("cls")
#define fs first
#define sc second
#define PC(x) putchar(x)
#define GC(x) x=getchar()
#define Endl PC('\n')
#define SF scanf
#define PF printf
inline int Read()
{
    int X=0,w=0;char ch=0;while(!isdigit(ch)){w|=ch=='-';ch=getchar();}while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
	return w?-X:X;
}
inline void Write(int x){if(x<0)putchar('-'),x=-x;if(x>9)Write(x/10);putchar(x%10+'0');}
inline LL powmod(LL a,LL b){LL RES=1;a%=MOD;assert(b>=0);for(;b;b>>=1){if(b&1)RES=RES*a%MOD;a=a*a%MOD;}return RES%MOD;}
inline LL gcdll(LL a,LL b){return b?gcdll(b,a%b):a;}
const int dx[]={1,1,2,2,-1,-1,-2,-2};
const int dy[]={2,-2,1,-1,2,-2,1,-1};
/************************************************************Begin************************************************************/
const int maxn=10;
int n,cnt[maxn][maxn];
bool vis[maxn][maxn];
pair<int,int> pre[maxn][maxn];
vector<pair<int,int> > v;
inline bool ok()
{
	rep(i,n) rep(j,n) if(!vis[i][j]) return 0;
	return 1;
}
inline void print(int x,int y)
{
	if(pre[x][y].fs!=-1) print(pre[x][y].fs,pre[x][y].sc);
	PF("%c%c\n",char(x+'a'),char(y+'1'));
}
inline bool cmp(pair<int,int> x,pair<int,int> y)
{
	return cnt[x.fs][x.sc]<cnt[y.fs][y.sc];
}
inline void dfs(int x,int y)
{
	vis[x][y]=1;
	if(ok())
	{
		print(x,y);
		exit(0);
	}
	vector<pair<int,int> > w;w.clear();
	rep(i,8)
	{
		int cx=x+dx[i],cy=y+dy[i];
		if(cx>=0&&cx<n&&cy>=0&&cy<n&&!vis[cx][cy]) w.push_back({cx,cy});
	}
	sort(All(w),cmp);
	repV(i,w)
	{
		int cx=i.fs,cy=i.sc;
		pre[cx][cy]={x,y};
		dfs(cx,cy);
	}
	vis[x][y]=0;
}
int main()
{
	SF("%d",&n);
	rep(i,n) rep(j,n)
	{
		v.push_back({i,j});
		rep(k,8)
		{
			int ci=i+dx[k],cj=j+dy[k];
			if(ci>=0&&ci<n&&cj>=0&&cj<n) cnt[i][j]++;
		}
	}
	sort(All(v),cmp);
	repV(it,v)
	{
		int i=it.fs,j=it.sc;
		MC(vis);
		MC(pre);
		pre[i][j]={-1,-1};
		dfs(i,j);
	}
	PF("IMPOSSIBLE");
	return 0;
}
/*************************************************************End**************************************************************/