pku 2488 A Knight's Journey （搜索 DFS）

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28697		Accepted: 9822

Description

Background 

The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey 

around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board,
but it is still rectangular. Can you help this adventurous knight to make travel plans? 



Problem 

Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes
how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves
followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number. 

If no such path exist, you should output impossible on a single line.

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

#include <cstdio>
#include <cstdlib>
#include <stack>
#include <cstring>

using namespace std;

const int MAX = 9;

const int dirx[8]={-1,1,-2,2,-2,2,-1,1},diry[8]={-2,-2,-1,-1,1,1,2,2};

typedef struct Point{
    int x,y;
}point;

int p,q,n;
bool visit[MAX][MAX];
point pre[MAX][MAX];
bool mark;
stack<int> stx,sty;

void printPath(int x,int y){
    stx.push(x);
    sty.push(y);

    int tx,ty;

    tx = pre[x][y].x;
    ty = pre[x][y].y;

    while(tx!=-1){
        stx.push(tx);
        sty.push(ty);
        x = pre[tx][ty].x;
        y = pre[tx][ty].y;
        tx = x;
        ty = y;
    }

    while(!stx.empty()){
        printf("%c%d",sty.top()-1+'A',stx.top());
        stx.pop();
        sty.pop();
    }

    printf("\n\n");
}

void dfs(int x,int y,int len){

    if(mark)return;
    if(len==p*q){
        printPath(x,y);
        mark = true;
        return;
    }

    int i,tx,ty;

    for(i=0;i<8;++i){

        tx = x+dirx[i];
        ty = y+diry[i];
        if(tx<1 || tx>p || ty<1 || ty>q)continue;
        if(visit[tx][ty])continue;

        pre[tx][ty].x = x;
        pre[tx][ty].y = y;
        visit[tx][ty] = true;
        dfs(tx,ty,len+1);
        visit[tx][ty] = false;
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //(Author : **** iaccepted)

    int i;
    scanf("%d",&n);
    for(i=1;i<=n;++i){
        printf("Scenario #%d:\n",i);
        scanf("%d %d",&p,&q);
        memset(visit,0,sizeof(visit));
        mark = false;
        pre[1][1].x = -1;
        pre[1][1].y = -1;
        visit[1][1] = true;
        dfs(1,1,1);
        visit[1][1] = false;

        if(!mark){
            printf("impossible\n\n");
        }
    }
    return 0;
}

题目意思：象棋中的马在一张棋盘上是否能不反复的走全然部格子。假设能走完输出走的路径（以字典序），假设没有一种走法能达到这种目标，则输出impossible。

思路就是DFS 搜下去，当走过的格子数达到格子总数时就打印路径。所以要用一个数组记录每一个定点的前驱节点。