HDU 1010 Tempter of the Bone(深搜)

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 86037    Accepted Submission(s): 23462

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
        

Author ZHANG, Zheng  

Source ZJCPC2004  

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好久沒有做搜尋方面的題目，手賤做了一下，簡單的一個這樣的題都能錯幾遍，也是沒誰了。

     題意：一個n*m的矩陣，老鼠的起點在矩陣中的'S'上，終點在矩陣中的'D'，其中'X'是牆，老鼠不能通過，'.'是路但是隻能通過一次，過了一次之後就不能再走這個地方了，終點D在第K秒是打開，這就要求老鼠能夠在第K秒是正好到達D點，如果不能就輸出NO，可以的話就輸出YES.

     不過這個題做的話需要剪枝，如果不剪枝就會逾時。至于剪枝可以通過起點與終點的最短距離與要求的時間K同奇偶的特性來進行判斷。同奇偶性是指起點與終點是确定的，是以在不考慮中間有沒有牆的情況下有一個最短的距離，這個距離的奇偶性如果與要求的K的奇偶性不相同的話，那他絕對不可能到達。不信的話可以試試。

代碼：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

int n,m,k;
char map[9][9];
bool v[9][9];
int jx[] = {0,1,0,-1};
int jy[] = {1,0,-1,0};
int x2,y2;
int pf;
int p;

void DFS(int xx,int yy,int cnt)
{
    if(xx == x2 && yy ==y2 && cnt == k)
    {
        pf = 1;
        return ;
    }
    if(pf == 1)
    {
        return ;
    }
    for(int i=0;i<4;i++)
    {
        int fx = xx + jx[i];
        int fy = yy + jy[i];
        int ans = abs(x2 - fx) + abs(y2 - fy);
        if((1+cnt+ans<=k) && (1+cnt+ans)%2 == p && fx>=0 && fx<n && fy>=0 && fy<m && map[fx][fy]!='X' && v[fx][fy] == 0)
        {
            v[fx][fy] = 1;
            DFS(fx,fy,cnt+1);
            v[fx][fy] = 0;
        }
    }
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        memset(v,0,sizeof(v));
        if(n == 0 && m == 0 && k == 0)
        {
            break;
        }
        p = k%2;
        pf = 0;
        int x1,y1;
        for(int i=0;i<n;i++)
        {
            getchar();
            for(int j=0;j<m;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j] == 'S')
                {
                    x1 = i;
                    y1 = j;
                }
                else if(map[i][j] == 'D')
                {
                    x2 = i;
                    y2 = j;
                }
            }
        }
        getchar();
        int pans = abs(y2 - y1) + abs(x2 - x1);
        if(pans>k || pans%2!=p)
        {
            printf("NO\n");
            continue;
        }
        v[x1][y1] = 1;
        DFS(x1,y1,0);
        if(pf == 1)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}