poj 2488 深搜

来源：互联网发布：绕x轴顺时针旋转矩阵编辑：程序博客网时间：2024/05/21 08:58

链接http://poj.org/problem?id=2488

题意：按照字典序遍历整个棋盘，（规定了遍历的策略）因此不能使用广搜，必须使用深搜

ac code

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#include <math.h>#include <queue>#include <stack>#include <map>#define INF 123123123#define MAX_INDEX 26using namespace std;struct node{    char A;    int n;};node result[MAX_INDEX*MAX_INDEX];int result_index = 0;int direction[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};int visit[MAX_INDEX][MAX_INDEX];int p,q,already;int dfs(int x,int y);void debug_visit();int debug_times;int main(){    int n;    cin >> n;    for(int kk = 1; kk<=n ; kk++)    {        debug_times = 0;        for(int i=0; i<MAX_INDEX; i++)            for(int j=0; j<MAX_INDEX; j++)                visit[i][j] = 0;        cin >> q >> p;        already = 0;        result_index = 0;        int res = dfs(0,0);        printf("Scenario #%d:\n",kk);        if(res)        {            for(int i=0; i<result_index; i++)                printf("%c%d",result[i].A,result[i].n+1);            printf("\n\n");        }        else        {            printf("impossible\n\n");        }    }    return 0;}int dfs(int x,int y){    if(!visit[x][y])    {        visit[x][y] = 1;        result[result_index].A = x + 'A';        result[result_index].n = y;        result_index++;        //debug_visit();        if(result_index == p*q )            return 1;        for(int i=0; i<8; i++)        {            if((x+direction[i][0]>=0)&&(x+direction[i][0]<p)&&(y+direction[i][1]>=0)&&(y+direction[i][1]<q)&&(!visit[x+direction[i][0]][y+direction[i][1]]))            {                if(dfs(x+direction[i][0],y+direction[i][1]))                    return 1;            }        }        visit[x][y] = 0;        result_index--;        return 0;    }    return 0;}void debug_visit(){    printf("trytimes = %d\n",debug_times++);    for(int i=0; i<p; i++)    {        for(int j=0; j<q; j++)            printf("%d ",visit[i][j]);        printf("\n");    }}

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