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題目傳送門

題意:中文題面

分析:雙層BFS,之前寫過類似的題.總結坑點:

　　1.步數小于等于T都是YES　　2. 傳送門的另一側還是傳送門或者墻都會死　　3. 走到傳送門也需要一步

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#include <bits/stdc++.h> using namespace std;char maze[2][11][11]; int dx[4] = {-1, 1, 0, 0}; int dy[4] = {0, 0, -1, 1}; int n, m, tot; bool vis[2][11][11]; struct Point {int x, y, z, step;Point () {}Point (int x, int y, int z, int step) : x (x), y (y), z (z), step (step) {} }; Point s, e;bool check(int x, int y, int z) {if (x < 1 || x > n || y < 1 || y > m || vis[z][x][y] || maze[z][x][y] == '*') return false;else return true; }bool BFS(void) {memset (vis, false, sizeof (vis));int res = 0x3f3f3f3f;queue<Point> que; que.push (s);vis[s.z][s.x][s.y] = true;while (!que.empty ()) {Point u = que.front (); que.pop ();if (u.x == e.x && u.y == e.y && u.z == e.z) {res = min (res, u.step); continue;}for (int i=0; i<4; ++i) {int tx = u.x + dx[i], ty = u.y + dy[i], tz = u.z;if (!check (tx, ty, tz)) continue;if (maze[tz][tx][ty] == '#') {if (maze[1-tz][tx][ty] == '*' || maze[1-tz][tx][ty] == '#' || vis[1-tz][tx][ty]) continue;vis[1-tz][tx][ty] = true;que.push (Point (tx, ty, 1 - tz, u.step + 1));continue;}vis[tz][tx][ty] = true;que.push (Point (tx, ty, tz, u.step + 1));}}return res <= tot; }int main(void) {int T; scanf ("%d", &T);while (T--) {scanf ("%d%d%d", &n, &m, &tot);for (int k=0; k<2; ++k) {if (k == 1) getchar ();for (int i=1; i<=n; ++i) {scanf ("%s", maze[k][i] + 1);for (int j=1; j<=m; ++j) {if (maze[k][i][j] == 'S') {s = Point (i, j, k, 0);}else if (maze[k][i][j] == 'P') {e = Point (i, j, k, 0);}}}}if (BFS ()) puts ("YES");else puts ("NO");}return 0; }

　　

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轉載于:https://www.cnblogs.com/Running-Time/p/4982500.html

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