Codeforces 1089E - Easy Chess - [DFS+特判][2018-2019 ICPC, NEERC, Northern Eurasia Finals Problem E]

題目連接：https://codeforces.com/contest/1089/problem/E

Elma is learning chess figures.

She learned that a rook can move either horizontally or vertically. To enhance her understanding of rook movement Elma’s grandmother gave Elma an 8 × 8 chess board and asked her to find a way to move the rook from a1 to h8 making exactly n moves, so that all visited cells are different.

A visited cell is the initial cell a1 and each cell on which the rook lands after a move.

Input

The input contains a single integer n (2 ≤ n ≤ 63) — the desired number of moves.

Output

Output a space-separated list of n+ 1 visited cells in the order they are visited by the rook.

All cells must be different. The list should start with a1 and end with h8. A solution always exists.

 

題意：

給你 $8 \times 8$ 的棋盤，有一個棋子車在 $a1$ 位置，如今要通過 $n$ 步移動到達 $h8$ 位置。

要求不能第二次到達曾經到達過的格子，且必須剛好移動 $n$ 步，請給出一種移動方案。

 

題解：

上來先敲個DFS，萬一能過呢？

而後本地跑了一下，發如今 $2 \le n \le 56$ 的範圍內跑的飛快，剩下也沒幾種狀況了，乾脆特判了以後本身寫一個固定走法便可。

 

AC代碼：

#include<bits/stdc++.h>
#define mk(x,y) make_pair(x,y)
using namespace std;
typedef pair<int,int> pii;
int n;
bool vis[10][10],ok;
vector<pii> ans;
void dfs(int x,int y,int d)
{
    if(ok) return;
    if(d>n) return;
    if(d==n && x==8 && y==8) {ok=1;return;}
    for(int i=1;i<=8;i++)
    {
        if(!vis[i][y])
        {
            ans.push_back(mk(i,y)); vis[i][y]=1;
            dfs(i,y,d+1);
            if(ok) return;
            ans.pop_back(); vis[i][y]=0;
        }
        if(!vis[x][i])
        {
            ans.push_back(mk(x,i)); vis[x][i]=1;
            dfs(x,i,d+1);
            if(ok) return;
            ans.pop_back(); vis[x][i]=0;
        }
    }
}
int main()
{
    while(cin>>n)
    {
        if(n<=56)
        {
            ans.clear(); memset(vis,0,sizeof(vis));
            ans.push_back(mk(1,1)); vis[1][1]=1;
            ok=0; dfs(1,1,0);
            for(int i=0;i<ans.size();i++)
            {
                if(i>0) printf(" ");
                printf("%c%d",'a'+ans[i].first-1,ans[i].second);
            }
            printf("\n");
        }
        else
        {
            for(int r=1;r<=6;r++)
            {
                if(r%2==0) {
                    for(int c=8;c>=1;c--) printf("%c%d ",'a'+c-1,r);
                } else {
                    for(int c=1;c<=8;c++) printf("%c%d ",'a'+c-1,r);
                }
            }
            printf("a7 a8 b8 b7 c7 c8 d8 d7 ");
            switch(n)
            {
            case 57: printf("h7 h8\n"); break;
            case 58: printf("e7 h7 h8\n"); break;
            case 59: printf("e7 f7 h7 h8\n"); break;
            case 60: printf("e7 f7 g7 h7 h8\n"); break;
            case 61: printf("e7 e8 f8 f7 h7 h8\n"); break;
            case 62: printf("e7 e8 f8 f7 g7 h7 h8\n"); break;
            case 63: printf("e7 e8 f8 f7 h7 g7 g8 h8\n"); break;
            }
        }
    }
}