杭电ACM 1010 Tempter of the Bone
Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 120163 Accepted Submission(s): 32462
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.
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.
'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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AC（1）
#include
#include
#include
#include
#define N 1005
using namespace std;
int c, r, arrtime;
int sx, sy, ex, ey;
int x;
int flag; int to[4][2] = { { 0,-1 },{ 0,1 },{ -1,0 },{ 1,0 } };
struct
{
char ch;
}maze[N][N];
void dfs(int x, int y, int time)
{
if (flag) return;
if (x <= 0 || x>r || y <= 0 || y>c) return;
int ans;
ans = arrtime - time - abs(ex - x) - abs(ey - y);
if (ans & 1 || ans < 0) return;
if (x == ex&&y == ey&&time == arrtime) {
flag = 1; return;
}
for (int i = 0; i<4; i++)
{
if (maze[x + to[i][0]][y + to[i][1]].ch != 'X')
{
maze[x + to[i][0]][y + to[i][1]].ch = 'X';//走过的地方变为墙
dfs(x + to[i][0], y + to[i][1], time + 1);
maze[x + to[i][0]][y + to[i][1]].ch = '.';//迷宫还原，以便下次广搜
}
}
}
int main(void)
{
while (cin >> r >> c >> arrtime)
{
if (c == 0 && r == 0 && arrtime == 0)
break;
else
{
x = 0;
for (int i = 1; i <= r; i++)
{
for (int j = 1; j <= c; j++)
{
cin >> maze[i][j].ch;
if (maze[i][j].ch == 'S')
{
sx = i;
sy = j;
}
if (maze[i][j].ch == 'X')
{
x++;
}
if (maze[i][j].ch == 'D')
{
ex = i;
ey = j;
}
}
getchar();
}
if (r*c - x <= arrtime) { cout << "NO" << endl; }
else
{
flag = 0;
maze[sx][sy].ch = 'X';
dfs(sx, sy, 0);
if (flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
}
}
#include
#include
#include
#include
#define N 1005
using namespace std;
int c, r, arrtime;
int sx, sy, ex, ey;
int x;
int flag;
struct
{
char ch;
}maze[N][N];
void dfs(int x, int y, int time)
{
if (flag) return;
if (maze[x][y].ch == '.' || maze[x][y].ch == 'S')
maze[x][y].ch = 'X';
if (x == ex&&y == ey&&time == arrtime) {
flag = 1; return;
}
if (x <= 0 || x>r || y <= 0 || y>c) return;
int ans;
ans = arrtime - time - abs(ex - x) - abs(ey - y);
if (ans & 1 || ans < 0) return;
if (maze[x - 1][y].ch != 'X')
{
maze[x - 1][y].ch == 'X';
dfs(x - 1, y, time + 1);
maze[x - 1][y].ch = '.';
}
if (maze[x + 1][y].ch != 'X')
{
maze[x + 1][y].ch == 'X';
dfs(x + 1, y, time + 1);
maze[x +1][y].ch = '.';
}
if (maze[x ][y-1].ch != 'X')
{
maze[x ][y-1].ch == 'X';
dfs(x , y-1, time + 1);
maze[x][y-1].ch = '.';
}
if (maze[x][y +1].ch != 'X')
{
maze[x][y + 1].ch == 'X';
dfs(x, y + 1, time + 1);
maze[x][y + 1].ch = '.';
}
return;
}
int main(void)
{
while (cin >> r >> c >> arrtime)
{
if (c == 0 && r == 0 && arrtime == 0)
break;
else
{
x = 0;
for (int i = 1; i <= r; i++)
{
for (int j = 1; j <= c; j++)
{
cin >> maze[i][j].ch;
if (maze[i][j].ch == 'S')
{
sx = i;
sy = j;
}
if (maze[i][j].ch == 'X')
{
x++;
}
if (maze[i][j].ch == 'D')
{
ex = i;
ey = j;
}
}
getchar();
}
if (r*c - x <= arrtime) { cout << "NO" << endl; }
else
{
flag = 0;
dfs(sx, sy, 0);
if (flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
}
}
} t-[abs(ex-sx)+abs(ey-sy)] //若此算式结果为非偶数，则无法恰好在T步内达到