HDU 1010 Tempter of the Bone

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1010

题目类型 DFS

知识点 奇偶剪枝

不知道的同学可以了解下

http://baike.baidu.com/view/7789287.htm?fr=aladdin

思路 深搜标记每个走过的点， 然后再回溯！将每一条路径都搜出来

#include <iostream>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define maxn 50
int m, n, step;
struct Point
{
    int x, y;
}Ps, Pe;
char maps[maxn][maxn];
bool vis[maxn][maxn];
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}, OK;
void DFS(int x,int y,int k);
int main()
{
    while(scanf("%d%d%d",&m,&n,&step), m+n+step)
    {
        OK = 0;
        memset(vis,0,sizeof(vis));
        for(int i=0 ; i<m; i++)
        {
            scanf("%s",maps[i]);
            for(int j=0; j<n; j++)
            {
                if(maps[i][j] == 'S')
                    Ps.x = i, Ps.y = j;
                if(maps[i][j] == 'D')
                    Pe.x = i, Pe.y = j;
            }
        }
        int s = (Ps.x + Ps.y)%2 , e = (Pe.x + Pe.y)%2;
        if( (s+e)%2 == step%2 )
            DFS(Ps.x,Ps.y,step);

        if(OK)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}
void DFS(int x,int y,int k)
{
    if(OK)
        return ;
    if(x == Pe.x && x == Pe.y && k == 0)
    {
        OK = 1;
        return ;
    }
    int len = abs(Pe.x - x) + abs(Pe.y - y);

    if(len < step)
        return ;

    for(int i=0; i<4; i++)
    {
        int nx = x + dir[i][0];
        int ny = y + dir[i][1];
        if(nx >= 0 && nx < m && ny >=0 && ny < n && maps[nx][ny] != 'X' && !vis[nx][ny])
        {
            vis[nx][ny] = 1;
            DFS(nx,ny,k-1);
            vis[nx][ny] = 0;
        }
    }
}