具体题目及要求:
输入输出示例:
这是个经典的基础搜索问题,很快就用bfs写好了代码,但提交总是超时,后来才知道有奇偶剪枝这个判断,加上后,ac。
具体代码如下:
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
const int dir[4][2] = {{0, 1}, {0, -1},
{1, 0}, {-1, 0}};
int temp[4];
int flag = 0;
int m, n, t;
bool a[8][8];
char graph[10][10];
void dfs(int x, int y, int step, int t)
{
if (graph[x][y] == 'D' && step == t) {
flag = 1;
return;
}
if (step == t)
return;
if (flag == 1)
return;
for (int i = 0; i < 4; i++) {
int nx = x + dir[i][0];
int ny = y + dir[i][1];
if (nx < 0 || nx > n - 1 || ny < 0 || ny > m - 1 || a[nx][ny] == 1 || graph[nx][ny] == 'X')
continue;
a[x][y] = 1;
dfs(nx, ny, step + 1, t);
a[x][y] = 0;
}
}
int main()
{
while (cin >> n >> m >>t && t && n && m) {
memset(a, 0, sizeof(a));
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
cin >>graph[i][j];
if (graph[i][j] == 'S') {
temp[0] = i;
temp[1] = j;
}
if (graph[i][j] == 'D') {
temp[2] = i;
temp[3] = j;
}
}
int xx = abs(temp[2] - temp[0]);
int yy = abs(temp[3] - temp[1]);
if ((t - (xx + yy)) % 2 == 0) // 奇偶剪枝
dfs(temp[0], temp[1], 0, t);
if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
flag = 0;
}
return 0;
}