Tempter of the Bone hdu 1010 (奇偶剪枝）

Problem Description

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

'X': a block of wall, which the doggie cannot enter;
'S': the start point of the doggie;
'D': the Door; or
'.': an empty block.

The input is terminated with three 0's. This test case is not to be processed.

Output

For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

Sample Input

4 4 5 S.X. ..X. ..XD .... 3 4 5 S.X. ..X. ...D 0 0 0

Sample Output

NO

YES

 

题意：小狗从s走到d，必须恰好走s步，问能否走到。

常规dfs/bfs,由于限定了步数，考虑奇偶剪枝

int k=step-s-fabs(ex-x)-fabs(ey-y);
//step总步数，s当前步数,exey 终点坐标；
如果k值为奇数，则无法到达。
if(k&1) 剪掉

ac:代码

#include
#include 
using namespace std;
#define ss second
#define ff first
#define pb push_back
template 
void read(T &x)
{
    static char ch;
    bool neg = false;
    while(!isdigit(ch = getchar())) (ch == '-') && (neg = true);
    for(x = ch - '0'; isdigit(ch = getchar()); x = x * 10 + ch - '0');
    (neg) &&(x = -x);
}
template void read(T &head, U &... tail)
{
    read(head), read(tail...);
}
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 205;
const int mod = 1e9 + 7;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};

int n, m, step;
char mp[maxn][maxn];
bool vis[maxn][maxn];
bool flag;
int sx, sy, ex, ey;

void dfs(int x, int y, int s)
{
    if(flag||s>step)return ;
    if(x==ex&&y==ey&&s==step)
    {
        flag=true;
    }
    int k=step-s-fabs(ex-x)-fabs(ey-y);
    if(k&1) return ;
    vis[x][y]=true;
    int xx, yy;
    for(int i = 0; i < 4; i++)
    {
        xx = x + dx[i]; yy = y + dy[i];
        if(xx >= n || xx < 0 || yy < 0 || yy >= m) continue;
        if(!vis[xx][yy])
        {
            vis[xx][yy]=true;
            dfs(xx,yy,s+1);
            vis[xx][yy]=false;
        }
    }


}

void work()
{
    memset(vis, false, sizeof(vis));
    flag = false;
    for(int i = 0; i < n; i++)
    {
        scanf("%s",mp[i]);
        for(int j=0;j