SimonHu-real 2017-04-25 15:19 采纳率: 0%

# 求看一道HDU的题目 HDU1010

`````` #include <cstdio>
#include<iostream>
#include <algorithm>
#include <cmath>
#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 == 'X'&&time != 0) 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;

dfs(x - 1, y, time + 1);
if(maze[x - 1][y].ch != 'X')
maze[x - 1][y].ch = '.';

dfs(x + 1, y, time + 1);
if (maze[x + 1][y].ch != 'X')
maze[x + 1][y].ch = '.';

dfs(x, y - 1, time + 1);
if (maze[x][y-1].ch != 'X')
maze[x][y - 1].ch = '.';

dfs(x, y + 1, time + 1);
if (maze[x][y + 1].ch != 'X')
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;
}
}
}
}
``````

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

• 写回答

#### 2条回答默认 最新

• 战在春秋 2017-04-25 21:37
关注
``````        if (maze[x][y].ch == 'X'&&time != 0) return;
if (maze[x][y].ch == '.' || maze[x][y].ch == 'S')
maze[x][y].ch = 'X';
``````

前一个if分支中当ch == 'X'，会执行return。而下面的if分支赋值ch = 'X'后，并不执行return。
逻辑上是矛盾的。

本回答被题主选为最佳回答 , 对您是否有帮助呢?
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