POJ 2488 A Knight's Journey

/*
p 横
q 竖
顺序...
这题要按照字典序顺序搜索，深搜策略，判断成功的条件是走的步数等于格子的数目
*/

#include <stdio.h>
#include <string.h>
#define MAXN 27

int map[MAXN][MAXN];
int p, q, cas, ok, step;;
int pathX[MAXN], pathY[MAXN];
int dr[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};            //字典序顺序

int legal(int x, int y){
    if(x <= 0 || y <= 0 || x > p || y > q)
        return 0;
    return 1;
}
void DFS(int x, int y){
    if( ok )
        return;
    pathX[step] = y;
    pathY[step] = x;
    step++;    
    if(step == p * q){        
        ok = 1;
        return;
    }
    int d;
    map[x][y] = 1;
    for(d = 0; d < 8; ++d){
        int newx = x + dr[d][1], newy = y + dr[d][0];
        if(legal(newx, newy) && !map[newx][newy] ){
            DFS(newx, newy);
            step--;
        }    
    }
    map[x][y] = 0;
}
void preProcess(){
    printf("Scenario #%d:\n",++cas);
    memset(map, 0, sizeof(map));
    ok = 0;
    step = 0;
}
void printPath(){
    int i;
    for(i = 0; i < p * q; ++i){
        printf("%c%d",pathX[i] + 'A' - 1, pathY[i]);
    }
    printf("\n");
}
int main(){
    int n, i;
    scanf("%d",&n);
    for(i = 1; i <= n; ++i){
        preProcess();
        scanf("%d %d",&p, &q);    
        DFS(1, 1);
        if( ok ){
            printPath();
        } else {
            printf("impossible\n");
        }    
        if(i != n)
            printf("\n");    
    }
    return 0;
}