Ural 1298 Knight 题解

目录

• ural 1298 knight 题解

ural 1298 knight 题解

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**************************************************************/